Question

using the information below, δh =
enter answer
for: 5c(s) + 6h₂(g) → c₅h₁₂(l). enter your answer
rounded to three significant figures and without units.
c(s) + o₂(g) → co₂(g) δh = -393.5 kj
c₅h₁₂(l) + 8o₂(g) → 5co₂(g) + 6h₂o(g) δh = -3244.8 kj
2h₂(g) + o₂(g) → 2h₂o(g) δh = -483.5 kj

Explanation:

Step1: Identify target and given reactions

Target reaction: \( 5C(s) + 6H_2(g)
ightarrow C_5H_{12}(l) \)
Given reactions:

1. \( C_5H_{12}(l) + 8O_2(g)

ightarrow 5CO_2(g) + 6H_2O(g) \quad \Delta H = -3244.8 \, \text{kJ} \)

1. \( C(s) + O_2(g)

ightarrow CO_2(g) \quad \Delta H = -393.5 \, \text{kJ} \)

1. \( 2H_2(g) + O_2(g)

ightarrow 2H_2O(g) \quad \Delta H = -483.5 \, \text{kJ} \)

Step2: Manipulate given reactions

• Reverse reaction 1: \( 5CO_2(g) + 6H_2O(g)

ightarrow C_5H_{12}(l) + 8O_2(g) \quad \Delta H = +3244.8 \, \text{kJ} \)

• Multiply reaction 2 by 5: \( 5C(s) + 5O_2(g)

ightarrow 5CO_2(g) \quad \Delta H = 5 \times (-393.5) = -1967.5 \, \text{kJ} \)

• Multiply reaction 3 by 3: \( 6H_2(g) + 3O_2(g)

ightarrow 6H_2O(g) \quad \Delta H = 3 \times (-483.5) = -1450.5 \, \text{kJ} \)

Step3: Sum manipulated reactions

Add reversed reaction 1, multiplied reaction 2, and multiplied reaction 3:
\( (5CO_2 + 6H_2O) + (5C + 5O_2) + (6H_2 + 3O_2)
ightarrow C_5H_{12} + 8O_2 + 5CO_2 + 6H_2O \)
Simplify (cancel \( 5CO_2, 6H_2O, 8O_2 \)):
\( 5C(s) + 6H_2(g)
ightarrow C_5H_{12}(l) \)

Step4: Calculate \( \Delta H \)

Sum the \( \Delta H \) values:
\( \Delta H = 3244.8 + (-1967.5) + (-1450.5) \)
\( \Delta H = 3244.8 - 1967.5 - 1450.5 \)
\( \Delta H = 3244.8 - 3418 \)
\( \Delta H = -173.2 \) (rounded to three significant figures)

Answer:

-173