Write a relation between ΔG and Q and define the meaning of each term and answer the following : (a) Why a rea

NCERT Exemplar · Q54

Q. Write a relation between ΔG and Q and define the meaning of each term and answer the following :

(a) Why a reaction proceeds forward when Q < K and no net reaction occurs when Q = K.

(b) Explain the effect of increase in pressure in terms of reaction quotient Q. for the reaction : CO

(g) + 3H2

(g) ⇌ CH4

(g) + H2O (g)

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Concept understanding — Gibbs Free Energy from K

Gibbs Free Energy from $K$ — The Bridge Between Thermodynamics and Equilibrium

Imagine you're pushing a heavy box across a rough floor. You push hard, but the box barely moves. The potential to move is there — you're applying force — but the actual motion is tiny. That's the difference between thermodynamic spontaneity (the push) and equilibrium (where the box sits, barely budging).

Gibbs Free Energy ( $\Delta G$ ) tells you the push — whether a reaction can happen. The equilibrium constant $K$ tells you how far it actually goes before stopping. The equation that links them is one of the most powerful in chemistry:

\Delta G^\circ = -RT \ln K

Let's unpack this from the ground up.

Step 1: What is $\Delta G$ ?

Gibbs Free Energy change ( $\Delta G$ ) measures the maximum useful work a reaction can do at constant temperature and pressure. More practically:

If $\Delta G < 0$ : the reaction is spontaneous (it can happen on its own).
If $\Delta G > 0$ : the reaction is non-spontaneous (it needs energy input).
If $\Delta G = 0$ : the system is at equilibrium — no net change.

But here's the catch: $\Delta G$ depends on how much reactant and product you have at any moment. It's not a fixed number.

Step 2: Standard vs. Non-standard Conditions

Chemists define a standard state (pure substances at 1 bar, 1 M concentration for solutions, 25°C usually). Under those conditions, the free energy change is called $\Delta G^\circ$ (standard Gibbs free energy change).

But real reactions rarely start at standard conditions. So we have:

\Delta G = \Delta G^\circ + RT \ln Q

where $Q$ is the reaction quotient (ratio of products to reactants at that instant, raised to their stoichiometric coefficients).

Note

$R$ is the gas constant (8.314 J/mol·K), $T$ is temperature in Kelvin. The $\ln$ is natural log.

Step 3: At Equilibrium — The Key Insight

At equilibrium, the reaction has no net tendency to go forward or backward. That means:

\Delta G = 0

And the reaction quotient $Q$ becomes exactly the equilibrium constant $K$ .

So plug into the equation:

0 = \Delta G^\circ + RT \ln K

Rearrange:

\Delta G^\circ = -RT \ln K

That's it. This single equation connects a thermodynamic property ( $\Delta G^\circ$ ) with a concentration-based constant ( $K$ ).

Step 4: What This Tells You

$\Delta G^\circ$ value	$K$ value	Meaning
Negative ( $<0$ )	$K > 1$	Products favoured at equilibrium
Zero ( $=0$ )	$K = 1$	Equal amounts at equilibrium
Positive ( $>0$ )	$K < 1$	Reactants favoured at equilibrium

Watch out

A negative $\Delta G^\circ$ does not mean the reaction is fast — only that it's thermodynamically favourable. Kinetics (activation energy) is a separate story.

Step 5: A Concrete Example

Consider the reaction: $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$

At 25°C, $\Delta G^\circ = -33.3 \text{ kJ/mol}$ . Using $R = 8.314 \text{ J/mol·K}$ :

$-33,300 = -(8.314)(298) \ln K$ …

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The relation $\Delta G = \Delta G^\circ + RT \ln Q$ connects the free energy change to the reaction quotient; a reaction proceeds forward when $Q < K$ because $\Delta G < 0$ , and at equilibrium $Q = K$ gives $\Delta G = 0$ . Increasing pressure on the given reaction shifts it forward because $Q$ temporarily drops below $K$ .

The Fundamental Relation

The connection between spontaneity and composition is captured by

\Delta G = \Delta G^\circ + RT \ln Q

where:

$\Delta G$ is the Gibbs free energy change under the current, non-standard conditions — it tells us whether the reaction will proceed forward ( $\Delta G < 0$ ), backward ( $\Delta G > 0$ ), or sit at equilibrium ( $\Delta G = 0$ ).
$\Delta G^\circ$ is the standard Gibbs free energy change, measured when all species are at unit activity (1 bar for gases, 1 M for solutions). It is a fixed number for a given reaction at a given temperature.
$R$ is the universal gas constant ( $8.314 \, \text{J mol}^{-1} \text{K}^{-1}$ ).
$T$ is the absolute temperature in kelvin.
$Q$ is the reaction quotient, the ratio of product activities to reactant activities raised to their stoichiometric coefficients, evaluated at any moment:

Q = \frac{[\text{products}]}{[\text{reactants}]}

At equilibrium, $\Delta G = 0$ and $Q$ becomes the equilibrium constant $K$ , so

0 = \Delta G^\circ + RT \ln K \quad \Rightarrow \quad \Delta G^\circ = -RT \ln K

Substituting this back into the first equation gives an alternative form:

\Delta G = RT \ln \frac{Q}{K}

This version makes the role of $Q$ and $K$ transparent.

(a) Why $Q < K$ drives the reaction forward, and $Q = K$ means equilibrium

The sign of $\Delta G$ determines direction. From $\Delta G = RT \ln \frac{Q}{K}$ :

When $Q < K$ : The ratio $\frac{Q}{K} < 1$ , so $\ln \frac{Q}{K} < 0$ , which makes $\Delta G < 0$ . A negative free energy change means the forward reaction is spontaneous — the system can lower its free energy by converting reactants into products. The reaction proceeds forward until $Q$ rises to equal $K$ .
When $Q = K$ : Now $\frac{Q}{K} = 1$ , so $\ln 1 = 0$ and $\Delta G = 0$ . The system is at equilibrium. The forward and reverse rates are equal, and there is no net change in composition. The free energy is at a minimum with respect to the reaction coordinate; any shift in either direction would increase $G$ .
When $Q > K$ : The ratio $\frac{Q}{K} > 1$ , so $\ln \frac{Q}{K} > 0$ and $\Delta G > 0$ . The reverse reaction is spontaneous; the system will convert products back into reactants until $Q$ falls to $K$ .

Tip

Think of $K$ as the target and $Q$ as the current position. The reaction always moves to close the gap: if you have too few products ( $Q < K$ ), make more; if you have too many ( $Q > K$ ), decompose some.

(b) Effect of increasing pressure on the reaction $\text{CO (g)} + 3\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_4\text{(g)} + \text{H}_2\text{O(g)}$

First, count moles of gas on each side:

Reactants: $1 + 3 = 4$ moles of gas
Products: $1 + 1 = 2$ moles of gas

The forward reaction reduces the total number of gas molecules.

What happens when we increase pressure?

For an ideal gas, partial pressure is proportional to molar concentration at constant temperature. The reaction quotient in terms of partial pressures is

Q_p = \frac{P_{\text{CH}_4} \cdot P_{\text{H}_2\text{O}}}{P_{\text{CO}} \cdot P_{\text{H}_2}^3}

When we suddenly compress the system (increase total pressure), every partial pressure scales up by the same factor, say $\alpha > 1$ . The new quotient becomes …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 17 on this concept.

TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
$\Delta_f H^\Theta$
(g) is respectively $-1216.3$ , $-553.5$ and $-393.5$ kJ mol $^{-1}$ . What is the value of $x$ (in kJ mol $^{-1}$ ) in the following reaction? BaCO $_3$ (s) $\xrightarrow{\Delta}$ BaO (s) + CO $_2$
(g) $- x$ (A) $-269.3$ (B) $269.3$ (C) $2163$ (D) $-2163$
›Reveal solutionSolution

$\Delta_r H^\Theta = \Delta_f H(BaO) + \Delta_f H(CO_2) - \Delta_f H(BaCO_3) = +269.3\text{ kJ mol}^{-1}$ , so $x = 269.3$ .

For $\text{BaCO}_3(s)\rightarrow \text{BaO}(s) + \text{CO}_2(g)$ :
$\Delta_r H^\Theta = \big[\Delta_f H(\text{BaO}) + \Delta_f H(\text{CO}_2)\big] - \Delta_f H(\text{BaCO}_3)$
$\Delta_r H^\Theta = \big[(-553.5) + (-393.5)\big] - (-1216.3) = -947.0 + 1216.3 = +269.3\ \text{kJ mol}^{-1}$ …
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TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
$369~^\circ \mathrm{C}$
›Reveal solutionSolution
The transition temperature of sulphur (rhombic to monoclinic) is a well‑known constant of 369 K (≈ 96 °C). The correct option is (B).

The question asks for the transition temperature of sulphur. This refers to the temperature at which the solid phase changes from the rhombic (α‑sulphur) to the monoclinic (β‑sulphur) form. This is a classic fact from solid‑state chemistry and phase diagrams. The key is to recall the value and its correct unit — many students confuse the Celsius and Kelvin scales.

Recall the known value

The rhombic‑to‑monoclinic transition for sulphur occurs at 96 °C under standard pressure. This is a standard datum in textbooks.

Convert to Kelvin

Since Kelvin = Celsius + 273.15, we have:

$96 + 273.15 \approx 369\ \text{K}$
(Often rounded to 369 K.)

Match with the options

(A) 369 °C → far too high (sulphur melts at ~115 °C).

(B) 369 K → matches the conversion. …
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TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
$Q. The difference between energy of an activated complex and the average energy of reactants is called? (A) Threshold energy (B) Lattice energy (C) Activation energy (D) Kinetic energy$
›Reveal solutionSolution
The question asks for the term that describes the energy difference between the activated complex (transition state) and the average energy of reactants. That term is activation energy, so the correct option is (C).

The key concept here is reaction energy profiles and the transition state theory. In a chemical reaction, reactants must overcome an energy barrier to form products. The highest-energy, unstable arrangement along the reaction path is called the activated complex (or transition state). The energy required to go from the average energy of the reactants up to that peak is precisely the activation energy — it’s the “hump” that must be surmounted.

Let’s break down why each option fits or doesn’t:

Understanding the definition

The difference between the energy of the activated complex and the average energy of the reactants is, by definition, the minimum energy that reactants must acquire to react. This is the activation energy ( $E_a$ ). It is always positive for an endothermic or exothermic reaction (except in barrierless reactions, where it’s zero).

Why not (A) Threshold energy?

Threshold energy is a related but distinct concept from collision theory: it is the minimum kinetic energy that colliding molecules must have for a reaction to occur. While it is numerically equal to activation energy in some simple models, the question specifically asks for the difference between the energy of the activated complex and the average energy of reactants — that is the textbook definition of activation energy, not threshold energy. Threshold energy is defined relative to the ground state of reactants, not their average energy.

Why not (B) Lattice energy? …
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TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
$^{-1}$
›Reveal solutionSolution
The enthalpy change for atomising CCl $_4$ (g) into gaseous atoms is found by reversing its formation from atoms and adding the atomisation enthalpies of C and Cl $_2$ . The result is 1304 kJ mol $^{-1}$ , option (A).

The reaction CCl $_4$ (g) $\rightarrow$ C(g) + 4Cl(g) is the complete atomisation of gaseous carbon tetrachloride. We are not measuring this directly — we build it from known thermodynamic data using Hess’s law.

The key idea: the standard enthalpy of formation $\Delta_f H^\Theta$ of a compound is the enthalpy change when it is made from its elements in their standard states. For CCl $_4$ (l), the standard state of carbon is graphite (C(s)) and of chlorine is Cl $_2$ (g). But our target starts from CCl $_4$ (g) and ends with gaseous atoms. So we need to:

start with CCl $_4$ (g),

break it down into its elements in standard states (reverse of formation),

then atomise those elements into gaseous atoms.

We are given $\Delta_{\text{vap}}H^\Theta$ for CCl $_4$ — that tells us the liquid-to-gas step, because the formation enthalpy given is for the liquid, not the gas. Let’s work through it.

Convert liquid CCl $_4$ to gaseous CCl $_4$

The given $\Delta_f H^\Theta(\text{CCl}_4) = -135.5$ kJ mol $^{-1}$ is for the liquid. Our target reaction starts with the gas, so we must first vaporise it:

$\text{CCl}_4(l) \rightarrow \text{CCl}_4(g)$ , $\Delta H = +30.5$ kJ mol $^{-1}$ (endothermic).

Reverse the formation of CCl $_4$ (g) from its elements

The formation reaction is:

$\text{C(s)} + 2\text{Cl}_2(g) \rightarrow \text{CCl}_4(l)$ , $\Delta H = -135.5$ kJ mol $^{-1}$ .

But we need the gas, so combine step 1:

$\text{C(s)} + 2\text{Cl}_2(g) \rightarrow \text{CCl}_4(g)$ , $\Delta H = -135.5 + 30.5 = -105.0$ kJ mol $^{-1}$ .

Reversing this gives:

$\text{CCl}_4(g) \rightarrow \text{C(s)} + 2\text{Cl}_2(g)$ , $\Delta H = +105.0$ kJ mol $^{-1}$ .

Atomise carbon

$\text{C(s)} \rightarrow \text{C(g)}$ , $\Delta_{\text{a}}H^\Theta(\text{C}) = 715.0$ kJ mol $^{-1}$ .

Atomise chlorine …
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TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
$x \times 10^5 \ \mathrm{ms}^{-1}$
›Reveal solutionSolution
The photoelectric effect gives the kinetic energy of emitted electrons as the photon energy minus the work function. Using the given wavelength and work function, the electron velocity is found to be about $4 \times 10^5$ m/s, so the nearest integer $x$ is 4.

Concept & Intuition

This is a classic photoelectric effect problem. When light of a certain frequency (or wavelength) hits a metal, each photon can transfer its energy to an electron. If the photon’s energy exceeds the metal’s work function (the minimum energy needed to free an electron), the excess becomes the electron’s kinetic energy. The kinetic energy then determines the electron’s speed via $K = \frac12 m v^2$ . The key is to convert all units consistently — here, photon energy is in joules (from wavelength) and work function is given in electronvolts, so we must convert eV to joules.

Step-by-step solution

Find the photon energy in joules The energy of a photon is $E = \frac{hc}{\lambda}$ , where $h = 6.63 \times 10^{-34} \ \text{J·s}$ , $c = 3 \times 10^8 \ \text{m/s}$ , $\lambda = 310 \ \text{nm} = 310 \times 10^{-9} \ \text{m}$ .

$E = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{310 \times 10^{-9}} = \frac{1.989 \times 10^{-25}}{3.1 \times 10^{-7}} \approx 6.416 \times 10^{-19} \ \text{J}.$

Convert the work function to joules Work function $\phi = 3.55 \ \text{eV}$ . Since $1 \ \text{eV} = 1.6 \times 10^{-19} \ \text{J}$ ,

$\phi = 3.55 \times 1.6 \times 10^{-19} = 5.68 \times 10^{-19} \ \text{J}.$

Compute the kinetic energy of the photoelectron By Einstein’s photoelectric equation:

$K = E - \phi = (6.416 - 5.68) \times 10^{-19} = 0.736 \times 10^{-19} \ \text{J}.$

Relate kinetic energy to velocity $K = \frac12 m_e v^2$ , with $m_e = 9 \times 10^{-31} \ \text{kg}$ . $v^2 = \frac{2K}{m_e} = \frac{2 \times 0.736 \times 10^{-19}}{9 \times 10^{-31}} …$
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TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
$^+$
›Reveal solutionSolution
The key idea is to relate the equilibrium constant to the standard cell potential using the Nernst equation at equilibrium. The result is $E^\Theta_{\text{cell}} = 0.45\ \text{V}$ , which corresponds to option (A).

We are given a reaction between a metal M and silver ions, with an enormous equilibrium constant ( $10^{15}$ ). That tells us the reaction strongly favors products — so the standard cell potential must be positive and fairly large. The bridge between $K$ and $E^\Theta$ is the Nernst equation at equilibrium, where the cell potential is zero and the reaction quotient equals $K$ .

Recall the fundamental relation At equilibrium, the Nernst equation becomes:

$E^\Theta_{\text{cell}} = \frac{0.0591}{n} \log K \quad (\text{at 298 K, using } \frac{2.303RT}{F} = 0.0591\ \text{V})$
But here they give $\frac{2.303RT}{F} = 0.06\ \text{V}$ (a rounded value), so we use:
$E^\Theta_{\text{cell}} = \frac{0.06}{n} \log K$
where $n$ is the number of moles of electrons transferred in the balanced reaction.

Determine $n$ The reaction is:

$\text{M(s)} + 2\text{Ag}^+(aq) \rightarrow \text{M}^{2+}(aq) + 2\text{Ag(s)}$
Each Ag⁺ gains one electron to become Ag, and two Ag⁺ ions are reduced, so total electrons transferred = 2. Also, M loses two electrons to become M²⁺. Hence $n = 2$ .

Plug in the numbers
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TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
$-4$
›Reveal solutionSolution
For an irreversible expansion against a constant external pressure, the work done by the gas is calculated using the external pressure and the change in volume. The work done is $\boxed{-4 \text{ atm L}}$ .

When a gas expands or contracts, it does work on its surroundings or has work done on it by the surroundings. The amount of work done depends on the path taken by the process.

For an irreversible process, especially when the expansion or compression occurs against a constant external pressure, the work done is determined by this external pressure, not the internal pressure of the gas. This is because the system (the gas) is not in equilibrium with its surroundings throughout the process. The external pressure is the opposing force that the gas must overcome to expand.
$W$
Here, the gas is expanding, so we expect the work done by the gas to be negative.

Identify the given values:

Initial volume, $V_1 = 10 \text{ L}$

Final volume, $V_2 = 12 \text{ L}$

External pressure, $P_{ext} = 2 \text{ atm}$

The process is irreversible.

Calculate the change in volume ( $\Delta V$ ):

The change in volume is the final volume minus the initial volume.

$\Delta V = V_2 - V_1 = 12 \text{ L} - 10 \text{ L} = 2 \text{ L}$

Apply the formula for work done in an irreversible process: …
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TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
$^+$
›Reveal solutionSolution
The key is linking the equilibrium constant to the standard Gibbs energy via $\Delta_r G^\Theta = -RT \ln K$ , and then using the given relation $RT/F \times 2.303 = 0.06$ V to compute the numerical value. The answer is -86.85 kJ mol⁻¹, which is option (C).

The problem gives you the equilibrium constant $K = 10^{15}$ for a redox reaction and asks for $\Delta_r G^\Theta$ in kJ mol⁻¹. The direct thermodynamic link is the equation $\Delta_r G^\Theta = -RT \ln K$ . But here they’ve supplied a specific value for $\frac{RT}{F} \times 2.303 = 0.06$ V, which is a hint that you can avoid plugging in $R$ and $T$ separately — just use this conversion factor.

Notice that $\ln K = 2.303 \log_{10} K$ , so $\Delta_r G^\Theta = -RT \times 2.303 \log_{10} K$ . And since $\frac{RT}{F} \times 2.303 = 0.06$ V, you have $RT \times 2.303 = 0.06 F$ . This lets you compute $\Delta_r G^\Theta$ directly in joules, then convert to kJ.

Write the standard relation:

$\Delta_r G^\Theta = -RT \ln K = -RT \times 2.303 \log_{10} K$ .

Substitute $\log_{10} K = \log_{10} (10^{15}) = 15$ :

$\Delta_r G^\Theta = -RT \times 2.303 \times 15$ .

Use the given: $RT \times 2.303 = 0.06 \times F$ , where $F = 96500$ C mol⁻¹. So:

$\Delta_r G^\Theta = - (0.06 \times 96500) \times 15$ J mol⁻¹.

Compute stepwise:

$0.06 \times 96500 = 5790$ J mol⁻¹.

Then $5790 \times 15 = 86850$ J mol⁻¹. …
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TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
$_2$
(g) and CH $_3$ OH(l) respectively are $-393$ , $-286$ and $-726$ kJ mol $^{-1}$ . What is the standard enthalpy of formation of methanol? (A) $-726$ kJ mol $^{-1}$ (B) $-239$ kJ mol $^{-1}$ (C) $-96$ kJ mol $^{-1}$ (D) $+96$ kJ mol $^{-1}$
›Reveal solutionSolution
Use Hess’s law: the enthalpy of formation of methanol is found by combining the combustion enthalpies of its elements and the combustion enthalpy of methanol. The result is $-239\ \text{kJ mol}^{-1}$ , option (B).

The key idea here is Hess’s law: enthalpy change for a reaction is the same whether it happens directly or in steps. We want the enthalpy of formation of methanol, i.e., the enthalpy change for:
$\text{C(graphite)} + 2\,\text{H}_2(g) + \tfrac12\,\text{O}_2(g) \rightarrow \text{CH}_3\text{OH}(l)$
We are given combustion enthalpies, which are enthalpies of complete burning in oxygen. So we can think of building the formation reaction from combustion steps.

Write the target formation reaction

$\text{C(s)} + 2\,\text{H}_2(g) + \tfrac12\,\text{O}_2(g) \rightarrow \text{CH}_3\text{OH}(l) \quad \Delta H_f^\circ = ?$

Write the given combustion reactions

Combustion of C(graphite):

$\text{C(s)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H = -393\ \text{kJ/mol}$

Combustion of H₂:

$\text{H}_2(g) + \tfrac12\,\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H = -286\ \text{kJ/mol}$

Combustion of methanol:

$\text{CH}_3\text{OH}(l) + \tfrac32\,\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\,\text{H}_2\text{O}(l) \quad \Delta H = -726\ \text{kJ/mol}$

Reverse the methanol combustion — because we want methanol as a product, not a reactant. Reversing changes the sign:

$\text{CO}_2(g) + 2\,\text{H}_2\text{O}(l) \rightarrow \text{CH}_3\text{OH}(l) + \tfrac32\,\text{O}_2(g) \quad \Delta H = +726\ \text{kJ/mol}$

Add the combustion of C and 2× combustion of H₂ to the reversed methanol combustion, so that CO₂ and H₂O cancel.

C combustion:

$\text{C(s)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H = -393$

2 × H₂ combustion:

$2\,\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\,\text{H}_2\text{O}(l) \quad \Delta H = 2 \times (-286) = -572$

Reversed methanol combustion:
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TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
$_2$
›Reveal solutionSolution
The enthalpy of combustion of CO is found by applying Hess’s law: the combustion reaction is CO + ½O₂ → CO₂, and its enthalpy equals ΔH_f(CO₂) − ΔH_f(CO) = −393 − (−110) = −283 kJ mol⁻¹, so the correct option is (A).

The key idea is that enthalpy of combustion is just the enthalpy change when one mole of a substance burns completely in oxygen. For CO, the combustion reaction is:
$\text{CO(g)} + \tfrac12 \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$
We are given the standard enthalpies of formation:

ΔH_f°(CO) = −110 kJ mol⁻¹

ΔH_f°(CO₂) = −393 kJ mol⁻¹

Since enthalpy is a state function, we can use Hess’s law: the enthalpy change for a reaction equals the sum of the enthalpies of formation of products minus the sum for reactants.

Write the target reaction

$\text{CO(g)} + \tfrac12 \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$

Apply the formation enthalpy formula For any reaction:

$\Delta H_{\text{rxn}} = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$
Here, the only product is CO₂, and the only reactant with a non-zero formation enthalpy is CO (O₂ is an element in its standard state, so its ΔH_f° = 0).

Plug in the numbers

$\Delta H_{\text{comb}} = \Delta H_f^\circ(\text{CO}_2) - \Delta H_f^\circ(\text{CO})$
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TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
$27^\circ \mathrm{C}$
›Reveal solutionSolution
Osmotic pressure is a colligative property given by $\pi = iCRT$ . For a non-electrolyte like sucrose, $i = 1$ , so $\pi = CRT$ . At $27^\circ \mathrm{C}$ ( $300\,\mathrm{K}$ ) with $C = 0.5\,\mathrm{M}$ and $R = 0.083\,\mathrm{L\,bar\,mol^{-1}\,K^{-1}}$ , the osmotic pressure is $12.45\,\mathrm{bar}$ , which corresponds to option (A).

Osmotic pressure is one of the four colligative properties — properties that depend only on the number of solute particles, not on their identity. The key formula is $\pi = iCRT$ , where $i$ is the van't Hoff factor (number of particles per formula unit in solution). Sucrose is a non-electrolyte; it dissolves as intact molecules, so $i = 1$ . That means the formula simplifies directly to $\pi = CRT$ .

The temperature must be in Kelvin. $27^\circ \mathrm{C}$ converts to $T = 27 + 273 = 300\,\mathrm{K}$ . The concentration is given as $0.5\,\mathrm{M}$ , which means $0.5$ moles per litre. The gas constant $R$ is provided in units of $\mathrm{L\,bar\,mol^{-1}\,K^{-1}}$ , so the answer will come out directly in bar — no unit conversion needed.

Write the formula: $\pi = CRT$ .

Substitute the values: $C = 0.5$ , $R = 0.083$ , $T = 300$ .

Multiply: $\pi = 0.5 \times 0.083 \times 300$ .

First, $0.5 \times 300 = 150$ .

Then, $150 \times 0.083 = 12.45$ . …
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TG EAPCET 2022Set ap-2022-07-30-FN1 markMCQ
$K_c$
›Reveal solutionSolution

The symmetric equilibrium ( $K_c = 9$ , all start at 1) gives $x = 0.5$ , so $[A] = 0.5$ and $[C] = 1.5$ — option (A).

Idea. Equal moles on both sides mean the total moles are unchanged, so we work directly with concentrations. Let $x$ react.

ICE table. Start: $[A]_0=[B]_0=[C]_0=[D]_0 = 1$ M. Let $x$ of A and B react:
$[A] = 1-x,\quad [B] = 1-x,\quad [C] = 1+x,\quad [D] = 1+x.$
Equilibrium expression.
$K_c = \frac{[C][D]}{[A][B]} = \frac{(1+x)^2}{(1-x)^2} = 9.$
Solve. Take the positive square root (all concentrations positive for $0<x<1$ ):
$\frac{1+x}{1-x} = 3 \;\Rightarrow\; 1+x = 3(1-x) \;\Rightarrow\; 1+x = 3-3x \;\Rightarrow\; 4x = 2 \;\Rightarrow\; x = 0.5.$
Concentrations.

$[A] = 1 - 0.5 = 0.5\ \text{M}, \qquad [C] = 1 + 0.5 = 1.5\ \text{M}.$ …
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Q.Write a relation between ΔG and Q and define the meaning of each term and answer the following :

Concept understanding — Gibbs Free Energy from K

Gibbs Free Energy from $K$ — The Bridge Between Thermodynamics and Equilibrium

Step 1: What is $\Delta G$ ?

Step 2: Standard vs. Non-standard Conditions

Step 3: At Equilibrium — The Key Insight

Step 4: What This Tells You

Step 5: A Concrete Example

Concept: Equilibrium Constant and Reaction Quotient

The Fundamental Relation

(a) Why $Q < K$ drives the reaction forward, and $Q = K$ means equilibrium

(b) Effect of increasing pressure on the reaction $\text{CO (g)} + 3\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_4\text{(g)} + \text{H}_2\text{O(g)}$

What happens when we increase pressure?

Q.Write a relation between ΔG and Q and define the meaning of each term and answer the following :

The Fundamental Relation

(a) Why Q<KQ < KQ<K drives the reaction forward, and Q=KQ = KQ=K means equilibrium

(b) Effect of increasing pressure on the reaction CO (g)+3H2(g)⇌CH4(g)+H2O(g)\text{CO (g)} + 3\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_4\text{(g)} + \text{H}_2\text{O(g)}CO (g)+3H2​(g)⇌CH4​(g)+H2​O(g)

What happens when we increase pressure?

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(a) Why $Q < K$ drives the reaction forward, and $Q = K$ means equilibrium

(b) Effect of increasing pressure on the reaction $\text{CO (g)} + 3\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_4\text{(g)} + \text{H}_2\text{O(g)}$

Unlock everything free for 14 days