# Question 1556c

Jun 23, 2016

ΔH = "-208 kJ"

#### Explanation:

The reaction is

$\text{CH"_4 + "2Cl"_2 → "CH"_2"Cl"_2 + "2HCl}$

Let's count the number of bonds in each substance.

$\textcolor{w h i t e}{l} \text{CH"_4color(white)(mll) = "4 C-H bonds}$
$\textcolor{w h i t e}{l} \text{2Cl"_2 color(white)(mll)= "2 Cl-Cl bonds}$
$\textcolor{w h i t e}{l} \text{CH"_2"Cl"_2 = "2 C-H bonds + 2 C-Cl bonds}$
$\textcolor{w h i t e}{l} \text{2HCl" color(white)(ml)= "2 H-Cl bonds}$

So, we have

$\textcolor{w h i t e}{m m m l l} \text{CH"_4 + color(white)(m)"2Cl"_2color(white)(ll) → color(white)(mmll)"CH"_2"Cl"_2color(white)(mm) + color(white)(l)"2HCl}$
$\textcolor{w h i t e}{m m m l} \text{4C-H" + color(white)(l)"2Cl-Cl" → color(white)(ll)"(2C-H + 2C-Cl)" + color(white)(l)"2H-Cl}$
$D \text{/kJ:"color(white)(l) "4×413"color(white)(ml) "2×242"color(white)(mml) "(2×413 + 2×328)" color(white)(ml)"2×431}$
$D \text{/kJ:"color(white)(ll) 1652 +color(white)(m) 484 color(white)(mmmml)"(826 + 656)"color(white)(ml) + color(white)(m)"862}$
$D \text{/kJ:} \textcolor{w h i t e}{m m m} 2136 \textcolor{w h i t e}{m m m m m m m m m m m} 2344$

ΔH = ΣD_"reactants" -ΣD_"products" = "2136 kJ -2344 kJ" = "-208 kJ"#