KTUYSBeta
Log inSign up

Thermodynamics · JEE Preparation Chemistry

Calculate the enthalpy of formation of benzene (text{C}6text{H}6(text{l})) given that the heats of combustion of benzene, carbon, and hydrogen are -3…

Practice question from Thermodynamics (JEE Preparation Chemistry), with the correct answer and a step-by-step solution.

medium · MCQ · Q-1A5DCD

Calculate the enthalpy of formation of benzene (C6H6(l)\text{C}_6\text{H}_6(\text{l})) given that the heats of combustion of benzene, carbon, and hydrogen are −3267 kJ mol−1-3267\text{ kJ mol}^{-1}, −393.5 kJ mol−1-393.5\text{ kJ mol}^{-1}, and −285.8 kJ mol−1-285.8\text{ kJ mol}^{-1} respectively.

  1. A.+48.5 kJ mol−1^{-1}Correct answer
  2. B.-48.5 kJ mol−1^{-1}
  3. C.+522.6 kJ mol−1^{-1}
  4. D.-1122.4 kJ mol−1^{-1}

Solution

The combustion reactions are:\n1) C6H6+152O2→6CO2+3H2O,ΔH1=−3267\text{C}_6\text{H}_6 + \frac{15}{2}\text{O}_2 \rightarrow 6\text{CO}_2 + 3\text{H}_2\text{O}, \Delta H_1 = -3267\n2) C+O2→CO2,ΔH2=−393.5\text{C} + \text{O}_2 \rightarrow \text{CO}_2, \Delta H_2 = -393.5\n3) H2+12O2→H2O,ΔH3=−285.8\text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{H}_2\text{O}, \Delta H_3 = -285.8\n\nFormation of benzene: 6C+3H2→C6H66\text{C} + 3\text{H}_2 \rightarrow \text{C}_6\text{H}_6\nΔHf=6ΔH2+3ΔH3−ΔH1=6(−393.5)+3(−285.8)−(−3267)=−2361−857.4+3267=+48.5 kJ mol−1\Delta H_f = 6\Delta H_2 + 3\Delta H_3 - \Delta H_1 = 6(-393.5) + 3(-285.8) - (-3267) = -2361 - 857.4 + 3267 = +48.5\text{ kJ mol}^{-1}.

More from Thermodynamics

All of Thermodynamics

Practise more Chemistry

Sign up free for adaptive practice on this chapter, with instant feedback and hints on every question.

Start practising free

Free to start · No credit card needed