Question

Consider the following intermediate chemical equations.(IMAGE) -205.7 kJ -113.4 kJ -14.3 kJ 78.0 kJ

Answer 1

Answer:

Approximately $-205.7\; \rm kJ$.

Explanation:

This question can be solved using Hess's Law.

Start by considering: how can the first three reactions (with known $\Delta H$ values) be combined to produce the reaction $\rm CH_4\; (g) + 4\; \rm Cl_2\; (g) \to CCl_4\; (g) + 4\; HCl\; (g)$?

Here's one possible combination:

• Include the first reaction once, without inverting.
• Invert the second reaction and include it once.
• Include the third reaction after multiplying all its coefficients by two.

In other words, if $(1)$, $(2)$, and $(3)$ denote the three reactions with know $\Delta H$ values, respectively, then $1 \times (1) - 1 \times (2) + 2\times (3)$ will give the required reaction $\rm CH_4\; (g) + 4\; \rm Cl_2\; (g) \to CCl_4\; (g) + 4\; HCl\; (g)$.

By Hess's Law, the $\Delta H$ value of the reaction $\rm CH_4\; (g) + 4\; \rm Cl_2\; (g) \to CCl_4\; (g) + 4\; HCl\; (g)$ will thus be:

$\begin{aligned}&1 \times \Delta H_1 - 1\times \Delta H_2 + 2\times \Delta H_3\\ &= 1 \times 74.6\; \rm kJ - 1 \times 95.7\; \rm kJ +2 \times (-92.3\; \rm kJ) \\ &= -205.7\; \rm kJ\end{aligned}$.