Q.The enthalpy of formation of CH4, CO2 and H2O are -74.8 kJ/mol, -393.5 kJ/mol and -285.8 kJ/mol respectively. Find Enthalpy of combustion of CH4.
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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 (ΔH) is simply the difference between the enthalpy of the products and the enthalpy of the reactants:
ΔH=Hproducts−Hreactants
Since enthalpy is a state function, ΔH 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
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 ΔH for the reaction:
A→D
But you can't measure it directly. You do know the enthalpy changes for these two steps:
- A→B ΔH1=+50 kJ/mol
- B→D ΔH2=−30 kJ/mol
Hess's Law says:
ΔHoverall=ΔH1+ΔH2=(+50)+(−30)=+20 kJ/mol
The path from A to D via B gives the same ΔH as the direct path would. You never have to measure the direct path.
Think of Hess's Law as algebra with chemical equations. You can add, subtract, reverse, and multiply entire reactions (and their ΔH 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 ΔH for such reactions using data from simpler, measurable reactions.
A classic example: the formation of carbon monoxide from carbon and oxygen.
C(s)+21O2(g)→CO(g)
If you burn carbon in limited oxygen, you always get some CO2 mixed in. But you can find ΔH for this reaction using two known values:
- C(s)+O2(g)→CO2(g) ΔH=−393.5 kJ/mol
- CO(g)+21O2(g)→CO2(g) ΔH=−283.0 kJ/mol …
Using Hess's law, the enthalpy of combustion of methane equals the sum of the enthalpies of formation of the products (CO2 and 2 H2O) minus the enthalpy of formation of methan …
Using ΔHcomb = ΣΔHf(products) - ΣΔHf(reactants) for CH4 + 2O2 → CO2 + 2H2O gives ΔHcomb = -890.3 kJ/mol.
Combustion reaction:
CH4(g)+2O2(g)→CO2(g)+2H2O(l)
Given standard enthalpies of formation:
- ΔfH(CH4) = -74.8 kJ/mol
- ΔfH(CO2) = -393.5 kJ/mol
- ΔfH(H2O) = -285.8 kJ/mol
- ΔfH(O2) = 0 (element in its standard state)
Applying Hess's law (enthalpy of reaction = sum of ΔfH of products − sum of ΔfH of reactants, each multiplied by its stoichiometric coefficient):
ΔHcomb=[ΔfH(CO2)+2ΔfH(H2O)]−[ΔfH(CH4)+2ΔfH(O2)]
Substituting values:
…
- CBSE 2025Set ANNUAL1 markMCQQ.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 …
- CBSE 2024Set ANNUAL1 markMCQQ.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
…
- CBSE 2023Set ANNUAL1 markQ.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.
…
- CBSE 2022Set ANNUAL1 markQ.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 …
- CBSE 2018Set ANNUAL1 markQ.(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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