Find the value of enthalpy of formation of PC | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The enthalpy of formation of $PCl_5(s)$ is the enthalpy change for the reaction: $\frac{1}{4} P_{4(s)} + \frac{5}{2} Cl_{2(g)} 	o PCl_{5(s)}$.Ste

Vedclass · Accepted Answer

$-454.5 \ kJ$

Vedclass · Answer

(D) The enthalpy of formation of $PCl_5(s)$ is the enthalpy change for the reaction: $\frac{1}{4} P_{4(s)} + \frac{5}{2} Cl_{2(g)} 	o PCl_{5(s)}$.Step $1$: Divide the first equation by $2$:$\frac{1}{4} P_{4(s)} + \frac{3}{2} Cl_{2(g)} 	o PCl_3(\ell) ; \Delta H = \frac{-635}{2} \ kJ = -317.5 \ kJ$Step $2$: Add the second equation:$PCl_3(\ell) + Cl_{2(g)} 	o PCl_{5(s)} ; \Delta H = -137 \ kJ$Step $3$: Add the two resulting equations:$\frac{1}{4} P_{4(s)} + (\frac{3}{2} + 1) Cl_{2(g)} 	o PCl_{5(s)} ; \Delta H = -317.5 + (-137) = -454.5 \ kJ$.

Find the value of enthalpy of formation of $PCl_5(s)$ given the following thermochemical equations:
$1) \frac{1}{2} P_{4(s)} + 3Cl_{2(g)} \to 2PCl_3(\ell) ; \Delta H = -635 \ kJ$
$2) PCl_3(\ell) + Cl_{2(g)} \to PCl_{5(s)} ; \Delta H = -137 \ kJ$

Similar Questions

Which factor affects the heat of reaction in the Kirchhoff equation?

The equation $\frac{1}{2} H_2 + \frac{1}{2} Cl_2 \to HCl$ $(\Delta H_{298} = -22.060 \ kcal)$ means:

Given that $C_{(g)} + 4H_{(g)} \to CH_{4(g)}; \Delta H = -166 \ kJ$,the bond energy of the $C-H$ bond will be....$kJ/mole$.

Among the following given substances,the one with zero $\Delta_{f}H^{\circ}$ is

The heat of reaction does not depend upon

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Find the value of enthalpy of formation of $PCl_5(s)$ given the following thermochemical equations:$1) \frac{1}{2} P_{4(s)} + 3Cl_{2(g)} \to 2PCl_3(\ell) ; \Delta H = -635 \ kJ$$2) PCl_3(\ell) + Cl_{2(g)} \to PCl_{5(s)} ; \Delta H = -137 \ kJ$