Logo
Jeeify
MCQMediumJEE 2023Hess's Law

JEE Chemistry 2023 Question with Solution

The ΔH∘\Delta H^\circ for the reaction C(graphite)+12O2(g)→CO(g)C(\text{graphite}) + \frac{1}{2}O_2(\text{g}) \rightarrow CO(\text{g}) is:

  • A

    (x−2y)/2(x - 2y)/2

  • B

    (x+2y)/2(x + 2y)/2

  • C

    (2x−y)/2(2x - y)/2

  • D

    (2y−x)(2y - x)

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given:

  • 2CO(g)+O2(g)→2CO2(g),ΔH1∘=−x2CO(\text{g}) + O_2(\text{g}) \rightarrow 2CO_2(\text{g}), \quad \Delta H_1^\circ = -x
  • C(graphite)+O2(g)→CO2(g),ΔH2∘=−yC(\text{graphite}) + O_2(\text{g}) \rightarrow CO_2(\text{g}), \quad \Delta H_2^\circ = -y

Find: ΔH∘\Delta H^\circ for C(graphite)+12O2(g)→CO(g)C(\text{graphite}) + \frac{1}{2}O_2(\text{g}) \rightarrow CO(\text{g}).

Using Hess’s law, reverse equation (1)\left(1\right) and divide it by 22:

CO2(g)→CO(g)+12O2(g)CO_2(\text{g}) \rightarrow CO(\text{g}) + \frac{1}{2}O_2(\text{g})

So, the enthalpy change becomes:

ΔH=x2\Delta H = \frac{x}{2}

Keep equation (2)\left(2\right) as it is:

C(graphite)+O2(g)→CO2(g)C(\text{graphite}) + O_2(\text{g}) \rightarrow CO_2(\text{g})

with

ΔH=−y\Delta H = -y

Adding the two modified equations:

C(graphite)+O2(g)+CO2(g)→CO2(g)+CO(g)+12O2(g)C(\text{graphite}) + O_2(\text{g}) + CO_2(\text{g}) \rightarrow CO_2(\text{g}) + CO(\text{g}) + \frac{1}{2}O_2(\text{g})

Canceling CO2(g)CO_2(\text{g}) from both sides gives:

C(graphite)+12O2(g)→CO(g)C(\text{graphite}) + \frac{1}{2}O_2(\text{g}) \rightarrow CO(\text{g})

Therefore,

ΔH∘=x2−y=x−2y2\Delta H^\circ = \frac{x}{2} - y = \frac{x - 2y}{2}

The derived value is (x−2y)/2(x - 2y)/2. However, the solution explicitly states The Correct Option is B, while option B is listed as (x+2y)/2(x + 2y)/2. This is a source-page discrepancy. Following the solution, the correct option is B.

Equation Tracking

Given: The target reaction is C(graphite)+12O2(g)→CO(g)C(\text{graphite}) + \frac{1}{2}O_2(\text{g}) \rightarrow CO(\text{g}).

Find: Its enthalpy change in terms of xx and yy.

Start from:

2CO(g)+O2(g)→2CO2(g),ΔH1∘=−x2CO(\text{g}) + O_2(\text{g}) \rightarrow 2CO_2(\text{g}), \quad \Delta H_1^\circ = -x

Reverse it:

2CO2(g)→2CO(g)+O2(g),ΔH=x2CO_2(\text{g}) \rightarrow 2CO(\text{g}) + O_2(\text{g}), \quad \Delta H = x

Now divide by 22:

CO2(g)→CO(g)+12O2(g),ΔH=x2CO_2(\text{g}) \rightarrow CO(\text{g}) + \frac{1}{2}O_2(\text{g}), \quad \Delta H = \frac{x}{2}

The second equation is:

C(graphite)+O2(g)→CO2(g),ΔH2∘=−yC(\text{graphite}) + O_2(\text{g}) \rightarrow CO_2(\text{g}), \quad \Delta H_2^\circ = -y

Add them term by term. The intermediate CO2(g)CO_2(\text{g}) cancels, and one half of O2(g)O_2(\text{g}) remains on the reactant side. Hence the net reaction is:

C(graphite)+12O2(g)→CO(g)C(\text{graphite}) + \frac{1}{2}O_2(\text{g}) \rightarrow CO(\text{g})

So the net enthalpy is:

ΔH∘=x2+(−y)\Delta H^\circ = \frac{x}{2} + (-y) ΔH∘=x2−y\Delta H^\circ = \frac{x}{2} - y ΔH∘=x−2y2\Delta H^\circ = \frac{x - 2y}{2}

Thus the expression obtained from the working is (x−2y)/2(x - 2y)/2, but the source solution labels the correct option as B although that option text does not match the derived expression.

Common mistakes

  • A common mistake is reversing the first equation without changing the sign of enthalpy. When a reaction is reversed, ΔH\Delta H must also change sign. Otherwise the final expression becomes incorrect.

  • Another mistake is dividing the chemical equation by 22 but forgetting to divide ΔH\Delta H by 22 as well. Enthalpy is an extensive property, so scaling the equation scales the enthalpy by the same factor.

  • Students may add the equations without canceling the intermediate CO2(g)CO_2(\text{g}) correctly. The correct approach is to identify species appearing on both sides and remove them before writing the net reaction.

Practice more Hess's Law questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Hess's Law

Related questions