Calculate the enthalpy change at 298.15 K and 1 bar pressure for the following reaction, using $\Delta_{\mathrm{f}} H_{208}^{\circ}$ data in Appendix 7:
CH₄ (g) + Cl₂ (g) → CH₃Cl (g) + HCl (g)

Question

Calculate the enthalpy change at 298.15 K and 1 bar pressure for the following reaction, using $\Delta_{\mathrm{f}} H_{208}^{\circ}$ data in Appendix 7: CH₄ (g) + Cl₂ (g) → CH₃Cl (g) + HCl (g)

Quizplus · Accepted Answer

To calculate the enthalpy change for the reaction, we need to subtract the sum of the enthalpies of the reactants from the sum of the enthalpies of the products:
ΔH = ΣH(products) - ΣH(reactants)

Using the given ΔfH° data from Appendix 7, we have:

ΔH°f(CH13SOCl(g)) = -388.5 kJ/mol
ΔH°f(Cl2(g)) = 0 kJ/mol
ΔH°f(CHSbU14SbU0(g)) = -30.9 kJ/mol
ΔH°f(HCl(g)) = -92.3 kJ/mol

ΔH = [ΔH°f(CH13SOCl(g)) + ΔH°f(HCl(g))] - [ΔH°f(CHSbU14SbU0(g)) + ΔH°f(Cl2(g))]
ΔH = [-388.5 - 92.3] - [-30.9 + 0] = -365.9 kJ/mol

Note that the coefficients in the balanced equation are all 1, so we don't need to multiply any of the enthalpies by a coefficient. Also, the ΔH we calculated is negative, indicating an exothermic reaction. The answer closest to -365.9 kJ/mol is A, -101.2 kJ/mol.

Calculate the Enthalpy Change at 298 ΔfH208∘\Delta_{\mathrm{f}} H_{208}^{\circ}Δf​H208∘​ Data in Appendix 7: CH4 (G) + Cl2 (G) →

Calculate the Enthalpy Change at 298 $\Delta_{\mathrm{f}} H_{208}^{\circ}$ Data in Appendix 7:
CH₄ (G) + Cl₂ (G) →