(i) Define: (a) Isotonic solution (b) Molecularity of reaction. (ii) State and explain Hess's law of constant

Question 114 of 115

Q. (i) Define:

(a) Isotonic solution

(b) Molecularity of reaction.

(ii) State and explain Hess's law of constant heat summation.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2026Subjective· 4mImportance★★★★★

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Concept understanding — Hess's Law

Hess's Law: The Chemistry Shortcut

Imagine you're climbing a mountain. You can take a direct, steep trail straight to the summit, or you can take a longer, winding path that goes through a valley first. Either way, you start at the base camp and end at the same peak. The net change in your altitude is exactly the same, no matter which route you take. The altitude difference between base and summit is a property of those two points alone — it doesn't care about the path.

That's the core idea behind Hess's Law.

In chemistry, enthalpy ( $H$ ) is like altitude. It's a state function — its value depends only on the current state of the system (temperature, pressure, composition), not on how you got there. When a reaction happens, the change in enthalpy ( $\Delta H$ ) is simply the difference between the enthalpy of the products and the enthalpy of the reactants:

\Delta H = H_{\text{products}} - H_{\text{reactants}}

Since enthalpy is a state function, $\Delta H$ for a given reaction is fixed. It doesn't matter if the reaction occurs in one step or in a dozen steps — the total enthalpy change will be identical.

The Precise Statement

Important

Hess's Law of Constant Heat Summation: The total enthalpy change for a chemical reaction is the same, regardless of the number of steps or the pathway by which the reaction occurs, provided the initial and final conditions are the same.

In other words, if you can break a reaction into a series of intermediate steps whose enthalpy changes you know, you can simply add them up to find the enthalpy change for the overall reaction.

How It Works in Practice

Suppose you want to find $\Delta H$ for the reaction:

\text{A} \rightarrow \text{D}

But you can't measure it directly. You do know the enthalpy changes for these two steps:

$\text{A} \rightarrow \text{B}$ $\quad \Delta H_1 = +50 \text{ kJ/mol}$
$\text{B} \rightarrow \text{D}$ $\quad \Delta H_2 = -30 \text{ kJ/mol}$

Hess's Law says:

\Delta H_{\text{overall}} = \Delta H_1 + \Delta H_2 = (+50) + (-30) = +20 \text{ kJ/mol}

The path from A to D via B gives the same $\Delta H$ as the direct path would. You never have to measure the direct path.

Tip

Think of Hess's Law as algebra with chemical equations. You can add, subtract, reverse, and multiply entire reactions (and their $\Delta H$ values) just like algebraic equations, as long as you keep track of what cancels.

Why This Matters

Many reactions are impossible to measure directly in a calorimeter. Maybe they're too slow, too dangerous, or they produce unwanted side products. Hess's Law lets you calculate $\Delta H$ for such reactions using data from simpler, measurable reactions.

A classic example: the formation of carbon monoxide from carbon and oxygen.

\text{C(s)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CO(g)}

If you burn carbon in limited oxygen, you always get some $\text{CO}_2$ mixed in. But you can find $\Delta H$ for this reaction using two known values:

$\text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$ $\quad \Delta H = -393.5 \text{ kJ/mol}$
$\text{CO(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$ $\quad \Delta H = -283.0 \text{ kJ/mol}$ …

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Isotonic solutions share osmotic pressure; molecularity counts species in an elementary step; Hess's law reflects that enthalpy is a state function.

(i)(a) Isotonic solutions: Two solutions that have the same osmotic pressure at a given temperature are called isotonic solutions. When separated by a semipermeable membrane, there is no net flow of solvent between them. Example: 0.9% aqueous NaCl solution is isotonic with human blood plasma.

(i)(b) Molecularity of a reaction: The number of reactant molecules, atoms or ions that come together (collide) in an elementary (single-step) reaction is called its molecularity. It is always a whole number (1, 2 or 3), determined directly from the stoichiometry of that elementary step (unlike order, which is experimentally determined and can be fractional or zero).

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

CBSE 2025Set ANNUAL1 markMCQ
$Q. The enthalpies of combustion of methane, graphite and hydrogen at 298 K are -890.3, -393.5 and -285.8 in kJ/mol respectively. Enthalpy of formation of CH4(g) will be$
(a) -74.8 kJ mol^-1
(b) -52.27 kJ mol^-1
(c) +74.8 kJ mol^-1
(d) +52.26 kJ mol^-1
›Reveal solutionSolution

Combine the three given combustion enthalpies via Hess's law to build the target formation reaction C(graphite) + 2H2(g) -> CH4(g) — the sum works out to -74.8 kJ/mol.

We want the enthalpy of formation of methane:

Target: C(graphite) + 2H2(g) -> CH4(g), Delta_f H = ?

Given combustion reactions:

(1) CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l), Delta H1 = -890.3 kJ/mol

(2) C(graphite) + O2(g) -> CO2(g), Delta H2 = -393.5 kJ/mol

(3) H2(g) + 1/2 O2(g) -> H2O(l), Delta H3 = -285.8 kJ/mol

By Hess's law, construct the target equation from these three: Target = (2) + 2x(3) - (1)

Check the atoms cancel correctly:

(2): C + O2 -> CO2

2x(3): 2H2 + O2 -> 2H2O

Sum: C + 2H2 + 2O2 -> CO2 + 2H2O …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The enthalpies of combustion of methane, graphite and dihydrogen at 298 K are -890.3, -393.5 and -285.8 in kJ mol^-1 respectively. Enthalpy of formation of CH4(g) will be$
(a) -74.8 kJ mol^-1
(b) -52.27 kJ mol^-1
(c) +74.8 kJ mol^-1
(d) +52.26 kJ mol^-1
›Reveal solutionSolution

Using Hess's law, the enthalpy of formation of methane equals the combustion enthalpy of carbon plus twice that of hydrogen, minus the combustion enthalpy of methane itself.

Target formation reaction: C(graphite) + 2 H2(g) -> CH4(g), delta Hf = ?

Given combustion enthalpies (delta Hc):

CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l), delta Hc(CH4) = -890.3 kJ/mol

C(graphite) + O2(g) -> CO2(g), delta Hc(C) = -393.5 kJ/mol

H2(g) + 1/2 O2(g) -> H2O(l), delta Hc(H2) = -285.8 kJ/mol

…

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CBSE 2023Set ANNUAL1 mark
$Q. State Hess's Law of Constant Heat Summation.$
›Reveal solutionSolution

Hess's Law states that the total enthalpy change for a reaction is the same whether it occurs in a single step or in a series of steps, since enthalpy is a state function.

Statement: If a reaction takes place in several steps, the standard enthalpy change of the overall reaction is equal to the sum of the standard enthalpy changes of the individual steps, regardless of the path or number of steps taken to go from reactants to products.

…

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CBSE 2022Set ANNUAL1 mark
$Q. Write Hess Law.$
›Reveal solutionSolution

Hess's Law: total enthalpy change of a reaction is independent of the path taken, depending only on the initial and final states.

Because enthalpy (H) is a state function, delta-H for converting reactants into products has one fixed value, regardless of whether the reaction happens directly in one step or through a series of intermediate steps. If a reaction can be written as the sum of two or more other reactions, its enthalpy change equals the sum of the enthalpy changes of those reactions. This principle lets chemists calculate delta-H for reactions that are hard to measure directly, by …

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CBSE 2018Set ANNUAL1 mark
$Q. (c) State Hess's law of constant heat summation.$
›Reveal solutionSolution

Hess's Law: total delta-H for a reaction is path-independent, depending only on initial and final states.

Step 1 -- statement: Hess's Law of Constant Heat Summation states that if a chemical reaction takes place in several steps, the standard enthalpy change (heat of reaction) for the overall reaction is equal to the sum of the standard enthalpy changes for the individual steps, and this total is the SAME regardless of the actual path or number of steps by which the reaction is carried out -- it depends only on the initial reactants and final products.

Step 2 -- basis: This follows because enthalpy (H) is a state function: delta-H for any process depends only on the initial and final states of the system, not on the route taken to get there. …

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