iszy1d45
contestada

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

Consider the following intermediate chemical equationsIMAGE 2057 kJ 1134 kJ 143 kJ 780 kJ class=

Respuesta :

Answer:

Approximately [tex]-205.7\; \rm kJ[/tex].

Explanation:

This question can be solved using Hess's Law.

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

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

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

[tex]\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}[/tex].