Question

Consider the following intermediate chemical equations.

Answer 1

Answer: 250 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.

$P_4(s)+6Cl_2\rightarrow 4PCl_3$  $\Delta H_1=-2439kJ$ (1)

$4PCl_5(g)\rightarrow P_4(s)+10Cl_2(g)$  $\Delta H_2=3438kJ$ (2)

Net chemical equation:

$PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g)$  $\Delta H=?$ (3)

Adding 1 and 2 we get,

$4PCl_5(g)\rightarrow 4PCl_3(g)+4Cl_2$ $\Delta H_3=\Delta H_1+\Delta H_2=-2439+3438=1000kJ$   (4)

Now dividing equation (4) by 4, we get

$PCl_5(g)\rightarrow PCl_3(g)+Cl_2$

$\Delta H=\frac{\Delta H_3}{4}=\frac{1000kJ}{4}=250kJ$   (4)