Question

Calculate the work done (in cal) when $0.425g$ of $H_2{O}_2$ decompose against a pressure of 1 atm at ${127}^{o}C$?
$2H_2{O}_2\left(\ell \right)\to O_2\left(g\right)+2H_{2}O\left(\ell \right)(R=2cal/Kmole)$

• A. $15$ Cal.
• B. $2$ Cal.
• C. $20$ Cal.
• D. $5$ Cal.

Answer 1

The correct option is D $5$ Cal.

The decomposition of hydrogen peroxide is represented by the following reaction.
$2H_2{O}_2\left(\ell \right)\to O_2\left(g\right)+2H_{2}O\left(\ell \right)$
The change in number of moles of gaseous species is $1-0=0$
Thus for 2 moles of hydrogen peroxide, $\Delta n$ is 1
The molar mass of hydrogen peroxide is 34 g/mol
0.425 g of hydrogen peroxide corresponds to $\frac{0.425}{34}$ moles.
For $\frac{0.425}{34}$ moles of hydrogen peroxide, the $\Delta n$ is $\frac{0.425}{34\times 2}\times 1$
Work done is the product of pressure and volume change.
$w=-P\Delta V$
$=-{\Delta n}_{g}RT=-\frac{0.425}{34\times 2}\times 2\times 400$
$w=-5Cal$.
Hence, the work done during decomposition of hydrogen peroxide is $-5Cal$.