Question

Calculate the work done (in J) when $4.5$ $g$ of $H_2{O}_2$ reacts against a pressure of $1.0$ $atm$ at ${25}^{o}C-$
$2H_2{O}_2\left(l\right)⟶O_2\left(g\right)+2H_{2}O\left(l\right)$

• A. $-1.6\times {10}^2$
• 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 $-1.6\times {10}^2$

Given, $△n_g=-\frac{4.5}{34\times 2}$

We know workdone,

$W=-P△V=-△n_{g}RT$

$W=-\frac{4.5}{34\times 2}\times 8.314\times 298$

$W=-1.63\times {10}^2$

Hence the correct option is [A].