**iii.** State Hess's law of constant heat summation. Illustrate with an example. State its applications.

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Q. iii. State Hess's law of constant heat summation. Illustrate with an example. State its applications.

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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:

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

Reverse the second equation (which flips the sign of $\Delta H$ ):

\text{CO}_2\text{(g)} \rightarrow \text{CO(g)} + \frac{1}{2}\text{O}_2\text{(g)} \quad \Delta H = +283.0 \text{ kJ/mol}

Now add it to the first equation. The $\text{CO}_2$ and $\text{O}_2$ cancel, leaving:

\text{C(s)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CO(g)} \quad \Delta H = -393.5 + 283.0 = -110.5 \text{ kJ/mol}

Watch out

When you reverse a reaction, always flip the sign of $\Delta H$ . When you multiply a reaction by a coefficient, multiply $\Delta H$ by the same coefficient. These are the two most common mistakes.

The Bottom Line

Hess's Law is not a new discovery — it's a direct consequence of enthalpy being a state function. The universe doesn't care about your reaction mechanism; it only cares about where you start and where you end. That's why you can calculate $\Delta H$ for any reaction by adding up known enthalpy changes from a carefully chosen pathway.

Hess's Law: The total enthalpy change of a reaction is independent of the pathway, because enthalpy is a state function.

Hess's Law is a numerical-heavy concept from the Thermodynamics chapter of NCERT/CBSE Class 11 Chemistry, and it shows up often in "Hess's Law numericals", "Hess's Law important questions", and "Hess's Law class 11 chemistry" searches because board exams, JEE Main, and NEET all test it through calculation-based problems.

Step 1. Hess's law of constant heat summation states that the overall enthalpy change of a reaction equals the sum of the enthalpy changes of the individual steps that combine to give that overall reaction.

Step 2. This follows directly from enthalpy being a STATE function -- the total change depends only on initial and final states, never on the path (number of steps) taken.

Step 3. Illustration: 2H2(g)+N2(g)→N2H4(g), ΔrH1=+95.4 kJ, followed by N2H4(g)+H2(g)→2NH3(g), ΔrH2=-187.6 kJ; adding these (N2H4 cancels) gives 3H2(g)+N2(g)→2NH3(g), ΔrH0=95.4-187.6=-92.2 kJ.

Step 4. Applications: Hess's law is used to find the enthalpy of reactions that cannot be measured directly (such as a compound's formation from its elements, e.g. Examples 4.14/4.15's Fe2O3 and SiC calculations), by combining several related, experimentally-known reactions instead.

✓Final answer

Hess's law: ΔrH(overall) = Σ ΔH(individual steps), a consequence of H being a state function; used to calculate otherwise-unmeasurable reaction enthalpies (e.g. NH3 synthesis: +95.4 + (-187.6) = -92.2 kJ).

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

Subtract (1): CH4 + 2O2 -> CO2 + 2H2O, i.e. reverse it: CO2 + 2H2O -> CH4 + 2O2

Adding [C + 2H2 + 2O2 -> CO2 + 2H2O] + [CO2 + 2H2O -> CH4 + 2O2] gives:

C + 2H2 -> CH4 (the O2 and CO2/H2O terms cancel) — exactly the target reaction.

So: Delta_f H(CH4) = Delta H2 + 2(Delta H3) - Delta H1

= (-393.5) + 2(-285.8) - (-890.3)

= -393.5 - 571.6 + 890.3

= -74.8 kJ/mol

✓Final answer
$(a) -74.8 kJ/mol.$
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

By Hess's law, combine: [C combustion] + 2 x [H2 combustion] - [CH4 combustion] cancels the O2 and gives exactly the formation reaction (the CO2 and H2O produced by C and H2 combustion are consumed back when CH4 combustion is reversed and subtracted).

delta Hf(CH4) = delta Hc(C) + 2 x delta Hc(H2) - delta Hc(CH4)

= (-393.5) + 2 x (-285.8) - (-890.3)

= -393.5 - 571.6 + 890.3

= -74.8 kJ/mol

✓Final answer
$(a) -74.8 kJ mol^-1.$
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.

This law follows directly from the fact that enthalpy (H) is a state function — its value depends only on the initial and final states of the system, not on the path taken between them. Hess's Law is very useful for calculating the enthalpy change of reactions that are difficult or impossible to carry out (or measure) directly, by combining the known enthalpy changes of other related reactions.

✓Final answer
$Hess's Law of Constant Heat Summation: the enthalpy change for a chemical reaction is the same whether the reaction occurs in one step or in several steps, since enthalpy is a state function and depends only on the initial and final states, not the path taken.$
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 combining the delta-H values of related, easily-measured reactions (Hess's Law of Constant Heat Summation).

✓Final answer
$Hess's Law: the enthalpy change of a reaction is the same whether it occurs in one step or several steps (enthalpy is a state function).$
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.

Step 3 -- utility: Hess's law allows the enthalpy change of a reaction that is difficult or impossible to measure directly to be calculated by combining the (measurable) enthalpy changes of other related reactions, using simple algebraic addition/subtraction of thermochemical equations.

✓Final answer
$Hess's Law: the total enthalpy change of a reaction is the same whether it takes place in one step or several steps -- it depends only on the initial and final states, not the path.$

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Q.iii. State Hess's law of constant heat summation. Illustrate with an example. State its applications.

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