Question

question 10 of 27
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) δh° = 90 kj/mol
½n₂(g)+o₂(g)→no₂(g) δh° = 34 kj/mol
a. 124 kj
b. -248 kj
c. -112 kj
d. 56 kj

Explanation:

Step1: Label the reactions

Let \(2NO(g)+O_2(g)
ightarrow2NO_2(g)\) be reaction (1), \(\frac{1}{2}N_2(g)+\frac{1}{2}O_2(g)
ightarrow NO(g)\) be reaction (2) with \(\Delta H_2 = 90\ kJ/mol\), and \(\frac{1}{2}N_2(g)+O_2(g)
ightarrow NO_2(g)\) be reaction (3) with \(\Delta H_3=34\ kJ/mol\).

Step2: Manipulate reactions

Multiply reaction (2) by 2: \(N_2(g) + O_2(g)
ightarrow2NO(g)\), \(\Delta H_{2 - new}=- 2\times90\ kJ/mol=-180\ kJ/mol\) (reverse - sign when reversing reaction). Keep reaction (3) as is: \(\frac{1}{2}N_2(g)+O_2(g)
ightarrow NO_2(g)\), \(\Delta H_3 = 34\ kJ/mol\). Multiply reaction (3) by 2: \(N_2(g)+2O_2(g)
ightarrow2NO_2(g)\), \(\Delta H_{3 - new}=2\times34\ kJ/mol = 68\ kJ/mol\).

Step3: Apply Hess's law

\(\Delta H_1=\Delta H_{3 - new}+\Delta H_{2 - new}\). Substitute the values: \(\Delta H_1=68+( - 180)=-112\ kJ/mol\).

Answer:

C. -112 kJ