ThermochemistryHard

Question

The difference between enthalpies of reaction at constant pressure and constant volume for the reaction

2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l) at 298 K in kJ is

Options

A.−7.43
B.+3.72
C.−3.72
D.+7.43

Solution

$\Delta H - \Delta E = \Delta n_{g}RT = (12 - 15) \times \frac{8.314}{1000} \times 298 = - 7432.7\text{ J}$

Create a free account to view solution

View Solution Free
Topic: Thermochemistry·Practice all Thermochemistry questions

More Thermochemistry Questions

Study the following thermochemical data.S + O2 → SO2: ΔH = −298.2 kJSO2 + ½ O2 → SO3: ΔH = −98.2 kJSO3 + H2O → H2SO4: ΔH...Study the following thermodynamic data given by E. H. P. Cordfunke, A. S. Booji and M. Y. Furkalionk.(i) DyCl3(s) → DyCl...Which of the following statement(s) is/are true?...The species which by definition has ZERO standard molar enthalpy of formation at 298 K is...For the given two processes(i) ½P4(s) + 3Cl2(g) → 2PCl3(l): ΔH = −635 kJ (ii) PCl3(l) + Cl2(g) → PCl5(s): ΔH = −137 kJth...