Question

question 9 of 30 what is the change in enthalpy of the first reaction below, given the enthalpies of the other two reactions?
2no(g) + o₂(g) → 2no₂(g)
½n₂(g) + ½o₂(g) → no(g) \tδh⁰ = 90 kj/mol
½n₂(g) + o₂(g) → no₂(g) \tδh⁰ = 34 kj/mol

a. 56 kj
b. -248 kj
c. -112 kj
d. 124 kj

Explanation:

Step1: Define the target and given reactions

Let the target reaction be \( R_1: 2NO(g) + O_2(g)
ightarrow 2NO_2(g) \)
Given reactions:
\( R_2: \frac{1}{2}N_2(g) + \frac{1}{2}O_2(g)
ightarrow NO(g) \quad \Delta H_2^0 = 90 \, \text{kJ/mol} \)
\( R_3: \frac{1}{2}N_2(g) + O_2(g)
ightarrow NO_2(g) \quad \Delta H_3^0 = 34 \, \text{kJ/mol} \)

Step2: Manipulate \( R_2 \) to reverse it and multiply by 2

Reverse \( R_2 \): \( NO(g)
ightarrow \frac{1}{2}N_2(g) + \frac{1}{2}O_2(g) \quad \Delta H = -90 \, \text{kJ/mol} \)
Multiply by 2: \( 2NO(g)
ightarrow N_2(g) + O_2(g) \quad \Delta H = 2\times(-90) = -180 \, \text{kJ/mol} \) (let's call this \( R_2' \))

Step3: Manipulate \( R_3 \) to multiply by 2

Multiply \( R_3 \) by 2: \( N_2(g) + 2O_2(g)
ightarrow 2NO_2(g) \quad \Delta H = 2\times34 = 68 \, \text{kJ/mol} \) (let's call this \( R_3' \))

Step4: Add \( R_2' \) and \( R_3' \) to get \( R_1 \)

\( R_2' + R_3' \):
\( 2NO(g) + N_2(g) + 2O_2(g)
ightarrow N_2(g) + O_2(g) + 2NO_2(g) \)
Simplify (cancel \( N_2(g) \) and subtract \( O_2(g) \) from both sides):
\( 2NO(g) + O_2(g)
ightarrow 2NO_2(g) \) (which is \( R_1 \))

Step5: Calculate \( \Delta H_1^0 \)

\( \Delta H_1^0 = \Delta H_{R_2'} + \Delta H_{R_3'} = -180 + 68 = -112 \, \text{kJ/mol} \)

Answer:

C. -112 kJ