Question

The heat of formation of HCl at 348 K from the following data, will be

0.5 H2(g) + 0.5 Cl2 (g) $\longrightarrow$HCl $\Delta$H⁰298 = – 22060 cal

The mean heat capacities over this temperature range are,

H2(g), CP = 6.82 calmol–1 K–1 ; Cl2 (g), Cp = 7.71 calmol–1 K–1 ; HCl (g), CP = 6.81 calmol–1 K–1

• A. – 20095 cal
• B. – 32758 cal
• C. – 37725 cal
• D. – 22083 cal

Answer 1

Answer: (D)

– 22083 cal

Answer 2

$\Delta$CP = CP (HCl) – $\frac{1}{2}$CP (H2) – $\frac{1}{2}$Cp (Cl2) = 6.81 – $\frac{1}{2}$ × 6.82 – $\frac{1}{2}$ × 7.71

= 6.81 – 3.41 – 3.855 = – 0.45

$\Delta$H348 = $\Delta$H298 + ($\Delta$CP) ($\Delta$T) = – 22060 + (– 0.45) × 50 = – 22082.5 cal