Question

$56g$ of iron reacts with $HCl$ in a closed vessel of fixed volume. Again, same amount of $Fe$ is made to react in an open beaker at ${27}^{o}C$. The work done (in cal) in both the cases will be respectively:

• A. $-600,-600$
• B. $-600,+600$
• C. $-300,-600$
• D. $0,-600$

Answer 1

The correct option is D $0,-600$

Solution:- (D) $0,-600$

In a closed vessel $\to$

When the volume is fixed,

$W=-P\Delta V$

$\because \Delta V=0$

$\Rightarrow W=0$

Hence for the reaction in closed vessel of fixed volume, work done will be zero.

In an open vesel $\to$

Molar mass of $Fe=56g$

Given mass of $Fe=56g$

As we know that,

$\text{No. of moles}=\frac{\text{Mass}}{\text{Molar mass}}$

Therefore, number of moles of $Fe$ used-

$n=\frac{56}{56}=1\text{moles}$

As we know that,

$w=-P\Delta V=-nRT[\because PV=nRT]$

Given $T=300K$

$\therefore w=-1\times 2\times 300=-600\text{cal}$

Hence for the reaction in an open vessel, the work done will be $-600\text{cal}$.