Question

consider the following intermediate chemical equations.
ch₄(g)→c(s)+2h₂(g) δh₁ = 74.8 kj
ccl₄(g)→c(s)+2cl₂(g) δh₂ = 95.7 kj
h₂(g)+cl₂(g)→2hcl(g) δh₃ = -92.3 kj
what is the enthalpy of the overall chemical reaction ch₄(g)+4cl₂(g)→ccl₄(g)+4hcl(g)?
-205.7 kj
-113.4 kj
-14.3 kj
78.0 kj

Explanation:

Step1: Identify the target reaction and given reactions

The target reaction is $CH_4(g)+4Cl_2(g)
ightarrow CCl_4(g) + 4HCl(g)$. The given reactions are:
Reaction 1: $CH_4(g)
ightarrow C(s)+2H_2(g)\quad\Delta H_1 = 74.8\ kJ$
Reaction 2: $CCl_4(g)
ightarrow C(s)+2Cl_2(g)\quad\Delta H_2=95.7\ kJ$
Reaction 3: $H_2(g)+Cl_2(g)
ightarrow 2HCl(g)\quad\Delta H_3=- 92.3\ kJ$

Step2: Manipulate the given reactions

Reverse Reaction 2 to get $C(s)+2Cl_2(g)
ightarrow CCl_4(g)\quad\Delta H_{2 - reversed}=- 95.7\ kJ$.
Multiply Reaction 3 by 2: $2H_2(g)+2Cl_2(g)
ightarrow 4HCl(g)\quad\Delta H_{3 - multiplied}=2\times(-92.3\ kJ)=-184.6\ kJ$

Step3: Add the manipulated reactions

Add Reaction 1, the reversed - Reaction 2 and the multiplied - Reaction 3:
$(CH_4(g)
ightarrow C(s)+2H_2(g))+(C(s)+2Cl_2(g)
ightarrow CCl_4(g))+(2H_2(g)+2Cl_2(g)
ightarrow 4HCl(g))$
The $C(s)$ and $2H_2(g)$ terms cancel out on the left - hand and right - hand sides.
The overall $\Delta H=\Delta H_1+\Delta H_{2 - reversed}+\Delta H_{3 - multiplied}$
$\Delta H = 74.8\ kJ-95.7\ kJ - 184.6\ kJ=-205.5\approx - 205.7\ kJ$

Answer:

$-205.7\ kJ$