Question

given the thermochemical equation:
1/2 n₂(g) + 1/2 o₂(g) → no(g) δh°rxn = +90.25 kj
what is the enthalpy change for the following reaction?
2 no(g) → n₂(g) + o₂(g) δh°rxn =?
+90.25 kj
-45.51 kj
-180.5 kj
-90.25 kj
+180.5 kj

Explanation:

Step1: Reverse the given reaction

When a reaction is reversed, the sign of $\Delta H$ changes.
The given reaction is \(\frac{1}{2}N_2(g)+\frac{1}{2}O_2(g)\to NO(g)\), \(\Delta H_{rxn}^o = + 90.25\space kJ\).
Reversing it gives \(NO(g)\to\frac{1}{2}N_2(g)+\frac{1}{2}O_2(g)\), \(\Delta H_{rxn}^o=-90.25\space kJ\)

Step2: Multiply the reversed reaction by 2

When a reaction is multiplied by a factor \(n\), the \(\Delta H\) is also multiplied by \(n\).
Multiplying \(NO(g)\to\frac{1}{2}N_2(g)+\frac{1}{2}O_2(g)\) by 2 gives \(2NO(g)\to N_2(g)+O_2(g)\)
\(\Delta H_{rxn}^o=2\times(- 90.25\space kJ)\)

Answer:

\(-180.5\space kJ\) (the third option)