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 kJP4(g) + 6 Cl2(g) → 4 PCl3(g) ΔH°rxn = -1207 kJ

Answer 1

Answer: The $\Delta H^o_{rxn}$ for the reaction is -1835 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$P_4(g)+10Cl_2(g)\rightarrow 4PCl_5(s)$      $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $PCl_5(s)\rightarrow PCl_3(g)+Cl_2(g)$    $\Delta H_1=157kJ$   ( × 4)

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

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[4\times (-\Delta H_1)]+[1\times \Delta H_2]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(4\times (-157))+(1\times (-1207))=-1835kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -1835 kJ.