Question

The enthalpy of formation of $N{H}_3$ is $-46KJ/mol$. The enhalpy change for the reaction $2N{H}_3⟶N_{2\left(g\right)}+3H_{2\left(g\right)}$ is:

• A. $+23KJ$
• B. $-92KJ$
• C. $+46KJ$
• D. $+184KJ$

Answer 1

The correct option is B $-92KJ$

In chemical reaction, $2N{H}_3\left(g\right)⟶N_2\left(g\right)+3H_2\left(g\right)$

$\Delta H_{reaction}=2\times \Delta H_f^o\left(N{H}_3\right)-\left[\Delta H_f^o\right(N_2)+3\Delta H_f^o(H_2\left)\right]$

$\Delta H_{reaction}=2(-46)-0$

$=-92kJ$

Hence the enthalpy of reaction is $-92kJ$ .