Question

When $280g$ of $N_2$ is converted into ammonia. $ZkJ$ of heat is evolved. WHat is the enthalpy of formation of ammonia?

• A. $20ZkJ$
• B. $-0.05ZkJ$
• C. $280ZkJ$
• D. $-2.8ZkJ$

Answer 1

The correct option is B $-0.05ZkJ$

Solution:- (B) $-0.05ZkJ$

${N_2}_{\left(g\right)}+3{H_2}_{\left(g\right)}⟶2{N{H}_3}_{\left(g\right)}$

Molecular weight of $N_2=28g$

$\therefore$ No. of moles in $280g$ of $N_2=\frac{280}{28}=10\text{moles}$

From the reaction,

$1$ mole of $N_2$ produces $2$ mole of $N{H}_3$.

$10$ mole of $N_2$ will produce $20$ mole of of $N{H}_3$.

As we know that enthalpy of formation is the amount of heat absorbed or released when $1$ mole of the product is formed.

Therefore,

Heat evolved when $10$ mole $N{H}_3$ formed $=ZkJ$

Heat evolved when $1$ mole $N{H}_3$ formed $=\frac{Z}{20}=0.05ZkJ$

Hence the enthalpy of formation of $N{H}_3$ is $-0.05ZkJ$ and negative sign indicates that the heat is evolved.