Question

At $25°\mathrm{C}$, $50$$\mathrm{g}$ of iron reacts with $\mathrm{HCl}$ to form ${\mathrm{FeCl}}_2$. The evolved ${\mathrm{H}}_2$ gas expands against a constant pressure of $1$ bar. The work done by the gas during this expansion is (round off to nearest integer) (given: $\mathrm{R}=8.14\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}$, assume ${\mathrm{H}}_2$ is an ideal gas, the atomic mass of $\mathrm{Fe}$is $55.85\mathrm{u}$) -

Answer 1

Step 1: Formula for work done for one mole of a gas- $\mathrm{W}=-\mathrm{P}∆\mathrm{V}\left(1\right)$

Where $\mathrm{P}$is the pressure and $∆\mathrm{V}$ is the change in volume during the process.

Substituting ideal gas equation - $\mathrm{PV}=\mathrm{nRT}$ in equation $\left(1\right)$, we get the formula for work done for one mole of a gas-

$\mathrm{W}=-∆{\mathrm{n}}_{\mathrm{g}}\mathrm{RT}$, where $∆{\mathrm{n}}_{\mathrm{g}}$ is the change in number of gaseous molecules, $\mathrm{R}$ is the gas constant, $\mathrm{T}$is the temperature.

Step 2: The reaction of iron with $\mathrm{HCl}$ to form ${\mathrm{FeCl}}_2$ is given below-

$$\begin{gathered} \mathrm{Fe}\left(\mathrm{s}\right)+2\mathrm{HCl}\left(\mathrm{aq}\right)\to {\mathrm{Fecl}}_2\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right) \\ \left(\mathrm{Iron}\right)(\mathrm{Hydrogen}\mathrm{chloride})(\mathrm{Ferrous}\mathrm{chloride})\left(\mathrm{dihydrogen}\right) \end{gathered}$$

$∆{\mathrm{n}}_{\mathrm{g}}$ for the reaction is $1$.

$\mathrm{R}=8.14\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}$

$\mathrm{T}=298\mathrm{K}$

Step 3: Work done for $\mathrm{n}$ number of moles-

$\mathrm{W}=-\mathrm{n}∆{\mathrm{n}}_{\mathrm{g}}\mathrm{RT}$

$\mathrm{n}=\frac{\mathrm{gven}\mathrm{mass}}{\mathrm{molar}\mathrm{mass}}$

$\mathrm{n}=\frac{50}{55.85}$

$\mathrm{W}=-1\times 8.14\times 298\times \frac{50}{55.85}$ $\mathrm{J}$

$\mathrm{W}=-2218$ $\mathrm{J}$

So, the work done by the gas during this expansion is (round off to nearest integer) is $-2218$ $\mathrm{J}$.