Question

Phosphorous pentachloride is used in the industrial preparation of many organic phosphorous compounds. Equation I shows its preparation from PCl3 and Cl2: (I) PCl3 (l) + Cl2(g)  PCl5(s) Use equation II and III to calculate ∆Hrxs of equation I: (II) P4 (s) + 6 Cl2 (g)  4 PCl3 (l) ∆H = 1280 KJ (III) P4 (s) + 10 Cl2 (g)  4 PCl5 (s) ∆H = 1774 KJ

Answer 1

Answer:

The enthalpy of the reaction is -123.5 kJ.

Explanation:

$P4 (s) + 6 Cl_2 (g)\rightarrow 4 PCl_3 (l) ,\Delta H_1 =-1280 kJ$..(1)

$P4 (s) + 10 Cl_2 (g)\rightarrow 4 PCl_5 (l) ,\Delta H_2 =-1774 kJ$..(2)

$PCl_3 (l) + Cl_2(g)\rightarrow PCl_5(s),\Delta H_{rxn}=x$...(3)

(2) - (1)

$4PCl_3 (l) + 4Cl_2(g)\rightarrow 4PCl_5(s),\Delta H_{rxn}=y$

Dividing equation by 4 we get (3)

$PCl_3 (l) + Cl_2(g)\rightarrow PCl_5(s),\Delta H_{rxn}=\frac{y}{4}$...(3)

$\Delta H_{rxn}=y=(-1774 kJ)-(-1280 kJ)=-494 kJ$

$\Delta H_{rxn}=x=\frac{y}{4}={-494 kJ}{4}=-123.5 kJ$

The enthalpy of the reaction is -123.5 kJ.