Question

Calculate the work done (in $J$) when $4.5g$ $H_2{O}_2$ reacts against a pressure of $1atm$ at ${25}^{o}C$
$2H_2{O}_2\left(l\right)⟶O_2\left(g\right)+H_{2}O\left(l\right)$

• A. $-1.49\times {10}^2$
• B. $4.5\times {10}^2$
• C. $3.2\times {10}^2$
• D. $-6.1\times {10}^2$

Answer 1

The correct option is A $-1.49\times {10}^2$

Solution:- (A) $-1.63\times {10}^{2}J$

Given mass of $H_2{O}_2=4.5g$

Molecular mass of $H_2{O}_2=34g$

No. of moles of $H_2{O}_2=\frac{4.5}{34}=0.132\text{mol}$

$2{H_2{O}_2}_{\left(l\right)}⟶{O_2}_{\left(g\right)}+2{H_{2}O}_{\left(l\right)}$

From the reaction-

$2$ moles of $H_2{O}_2$ produces $1$ mole of $O_2$, thus $0.132$ mole of $H_2{O}_2$ will produce $0.066$ mole of $O_2$.

$\Delta n_g=n_P-n_R=(0.066+0)-0=0.066\text{mol}$

$\therefore$ Change in volume $=\Delta V=\Delta n_g\times 22.4=0.066\times 22.4=1.48$

Pressure $=1atm\left[\text{Given}\right]$

As we know that,

Work done $\left(W\right)=-P\Delta V=-(1\times 1.48)=-1.48L-atm=-1.49\times {10}^{2}J$