Question

Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction: P4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°rxn = ? Given: PCl5(s) → PCl3(g) + Cl2(g) ΔH°rxn= +157 kJ P4(g) + 6 Cl2(g) → 4 PCl3(g) ΔH°rxn = -1207 kJ

Answer 1

Answer: -1835 kJ

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to Hess’s law, the chemical equation can be treated as algebraic expressions and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

$PCl_5(s)\rightarrow PCl_3(g)+Cl_2(g)$  $\Delta H_1=+157kJ$ (1)

$P_4(g)+6Cl_2\rightarrow 4PCl_3(g)$  $\Delta H_2=-1207J$ (2)

Net chemical equation:

$P_4(g)+10Cl_2(g)\rightarrow 4PCl_5(s)$  $\Delta H=?$ (3)

Multiplying equation (1) by 4,  and reversing we get

$4PCl_3(g)+4Cl_2(g)\rightarrow 4PCl_5(s)$  $\Delta H_4=4\times -157kJ=-628kJ$ (4)

Adding (2) and (4)

$P_4(g)+10Cl_2(g)\rightarrow 4PCl_5(s)$  $\Delta H_3=\Delta H_2+\Delta H_4=-1207kJ-628kJ=-1835kJ$  

Thus enthaply change for the reaction is -1835 kJ.