Question

What is the amount of heat released when 3.4 gm $N{H}_3\left(g\right)$ and 6.4 gm $O_2\left(g\right)$ react at constant temperature & pressure by the following equation?
$4N{H}_3\left(g\right)+5O_2\left(g\right)\to 4NO\left(g\right)+6H_{2}O\left(g\right).\Delta r{H}^o=-900kJ$

• A. 45 kJ
• B. 250 kJ
• C. 36 kJ
• D. None of these

Answer 1

The correct option is C 36 kJ

Solution:- (C) $36KJ$

Mol. wt. of ${N{H}_3}_{\left(g\right)}=17g$

Mol. wt. of ${O_2}_{\left(g\right)}=32g$

As we know that,

$\text{No. of moles}=\frac{\text{Wt.}}{\text{Mol. wt.}}$

Therefore,

No. of moles in $3.4gm$ of ${N{H}_3}_{\left(g\right)}=\frac{3.4}{17}=0.2\text{mol}$

No. of moles in $6.4gm$ of ${O_2}_{\left(g\right)}=\frac{6.4}{32}=0.2\text{mol}$

$4{N{H}_3}_{\left(g\right)}+5{O_2}_{\left(g\right)}⟶4{NO}_{\left(g\right)}+6{H_{2}O}_{\left(l\right)}$

According to reaction,

$5$ mole of ${O_2}_{\left(g\right)}$ required to rect with $4$ mole of ${N{H}_3}_{\left(g\right)}$

Therefore, $0.25$ mole of ${O_2}_{\left(g\right)}$ required to react with $0.2$ mole of ${N{H}_3}_{\left(g\right)}$.

But the given amount of ${O_2}_{\left(g\right)}$ is $0.2$ mole.

Now,

Amount of heat released when $5$ mole of $O_2$ react $=-900KJ$

$\therefore$ Amount of heat released when $0.25$ mole of ${O_2}_{\left(g\right)}$ react $=\frac{900}{5}\times 0.2=36KJ$