Thermodynamics Question 8
Question 8 - 31 January - Shift 2
Enthalpies of formation of $CCl_4(g), H_2 O(g), CO_2(g)$ and
$HCl(g)$ are $-105,-242,-394$ and $-92 kJ mol^{-1}$ respectively. The magnitude of enthalpy of the reaction given below is $kJ mol^{-1}$ (nearest integer)
$CCl_4(g)+2 H_2 O(g) \to CO_2(g)+4 HCl(g)$
Show Answer
Answer: (173)
Solution:
Formula: Enthalpy Change
$ \begin{aligned} & \Delta_r H=\sum H_p-\sum H_R \text{ mathongo matho } \\ & \quad=(-394+4 \times-92)-(-105+(2 \times-242)) \\ & \quad=-173 kJ / mol \end{aligned} $
