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NVAEasyJEE 2023Hess's Law

JEE Chemistry 2023 Question with Solution

Consider the following data:

Heat of combustion of H2(g)\mathrm{H_2(g)} = −241.8 kJ mol−1-241.8 \, \text{kJ mol}^{-1}

Heat of combustion of C(s)\mathrm{C(s)} = −393.5 kJ mol−1-393.5 \, \text{kJ mol}^{-1}

Heat of combustion of C2H5OH(l)\mathrm{C_2H_5OH(l)} = −1234.7 kJ mol−1-1234.7 \, \text{kJ mol}^{-1}

The heat of formation of C2H5OH(l)\mathrm{C_2H_5OH(l)} is (−)(-) _____ kJ mol−1\text{kJ mol}^{-1} (Nearest integer).

Answer

Correct answer:278

Step-by-step solution

Standard Method

Given:

  • Heat of combustion of H2(g)\mathrm{H_2(g)} is −241.8 kJ mol−1-241.8 \, \text{kJ mol}^{-1}
  • Heat of combustion of C(s)\mathrm{C(s)} is −393.5 kJ mol−1-393.5 \, \text{kJ mol}^{-1}
  • Heat of combustion of C2H5OH(l)\mathrm{C_2H_5OH(l)} is −1234.7 kJ mol−1-1234.7 \, \text{kJ mol}^{-1}

Find: The heat of formation of C2H5OH(l)\mathrm{C_2H_5OH(l)}.

Using Hess's Law Enthalpy Calculation, write the combustion equations shown in the solution:

2C(s)+2O2→2CO22\mathrm{C(s)} + 2\mathrm{O_2} \rightarrow 2\mathrm{CO_2} ΔH=−393.5×2=−787  kJ\Delta H = -393.5 \times 2 = -787 \; \text{kJ}3H2+32O2→3H2O3\mathrm{H_2} + \frac{3}{2}\mathrm{O_2} \rightarrow 3\mathrm{H_2O} ΔH=−241.5×3=−725.4  kJ\Delta H = -241.5 \times 3 = -725.4 \; \text{kJ}C2H5OH+3O2→2CO2+3H2O\mathrm{C_2H_5OH} + 3\mathrm{O_2} \rightarrow 2\mathrm{CO_2} + 3\mathrm{H_2O} ΔH=−1234.7  kJ\Delta H = -1234.7 \; \text{kJ}

Reverse the ethanol combustion equation:

3H2O+2CO2→C2H5OH+3O23\mathrm{H_2O} + 2\mathrm{CO_2} \rightarrow \mathrm{C_2H_5OH} + 3\mathrm{O_2} ΔH=+1234.7  kJ\Delta H = +1234.7 \; \text{kJ}

Now add the equations to obtain the target reaction:

2C(s)+3H2(g)+12O2→C2H5OH(l)2\mathrm{C(s)} + 3\mathrm{H_2(g)} + \frac{1}{2}\mathrm{O_2} \rightarrow \mathrm{C_2H_5OH(l)}

So,

ΔH=(−787)+(−725.4)+(1234.7)\Delta H = (-787) + (-725.4) + (1234.7)ΔH=−277.7  kJ\Delta H = -277.7 \; \text{kJ}ΔH≈−278  kJ\Delta H \approx -278 \; \text{kJ}

Therefore, the heat of formation is −278 kJ mol−1-278 \, \text{kJ mol}^{-1}, so the numerical answer is 278.

Direct Hess's law combination

Given: The combustion enthalpies of H2(g)\mathrm{H_2(g)}, C(s)\mathrm{C(s)}, and C2H5OH(l)\mathrm{C_2H_5OH(l)}.

Find: The heat of formation of C2H5OH(l)\mathrm{C_2H_5OH(l)}.

Use the idea that formation of ethanol can be obtained by adding the combustion of its elements and subtracting the combustion of ethanol itself.

Thus,

ΔHf=2(−393.5)+3(−241.8)−(−1234.7)\Delta H_f = 2(-393.5) + 3(-241.8) - (-1234.7)ΔHf=−787−725.4+1234.7\Delta H_f = -787 - 725.4 + 1234.7ΔHf=−277.7  kJ mol−1\Delta H_f = -277.7 \; \text{kJ mol}^{-1}ΔHf≈−278  kJ mol−1\Delta H_f \approx -278 \; \text{kJ mol}^{-1}

Therefore, the required numerical value is 278.

Common mistakes

  • Using the combustion enthalpy of ethanol with the wrong sign after reversing the reaction is a common mistake. When a reaction is reversed, the sign of ΔH\Delta H must also reverse. Use +1234.7 kJ mol−1+1234.7 \, \text{kJ mol}^{-1} for the reversed combustion equation.

  • Multiplying the combustion enthalpy of hydrogen or carbon by the wrong stoichiometric coefficient gives an incorrect result. The formation of C2H5OH\mathrm{C_2H_5OH} requires 22 carbon atoms and 33 molecules of H2\mathrm{H_2}, so use factors 22 and 33 respectively.

  • Reporting the final answer as −278-278 in the answer box is incorrect here because the question already states the value is (−)(-) _____. The blank requires only the magnitude, so enter 278.

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