Question

What is the work done when 50g of iron reacts with hydrochloric acid in an open beaker at ${25^0}C$ ?
A) $-22.12\;joule$
B) $+22.12\;joule$
C) $+2212.22\; joule$
D) $-2212.22 \;joule$

Answer 1

Work done by a body is defined as the energy required to displace one object from one place to another. In thermodynamics, the work done by expanding gases is stated as the integral of change in volume and pressure of the gas.

Complete answer:
Let us see the reaction between Iron and hydrochloric acid,
$F{e_{\left( s \right)}} + 2HCl{{\text{ }}_{\left( {aq} \right)}} \to FeC{l_{2{\text{ }}}}_{\left( {aq} \right)} + {\text{ }}{H_2}_{\left( g \right)}$
For every mole of Fe, one mole of H2 is produced.
We know that,
$Work{\text{ }}done = - P{\text{ }}\Delta V$
But according to the ideal gas law: PV= nRT
$Work{\text{ }}done = - P{\text{ }}\Delta V{\text{ = - nRT}}$
So, replacing $P{\text{ }}\Delta V{\text{ }}$ with nRT we get,
$\Rightarrow Work{\text{ }}done = - {\text{nRT}}$
Where,
n = number of moles
R = gas constant
T = temperature in Kelvin
First we need to find the number of moles of iron.
Molar mass of iron = 56g
Given mass of iron = 50g
\Rightarrow No.{\text{ of moles of Fe = }}\dfrac{{Molar{\text{ of Fe}}}}{{Given{\text{ mass of Fe}}}} \\\ {\text{ = }}\dfrac{{50}}{{56}} \\\ {\text{ = 0}}{\text{.893 moles}} \\\
\Rightarrow {\text{n = 0}}{\text{.893}} \\\ {\text{R = 8}}{\text{.314 J}}{{\text{K}}^{ - 1}}{\text{mol}}{{\text{e}}^{ - 1}}{\text{ }} \\\
To convert Celsius to kelvin- add 273 to the Celsius temperature.
$T{\text{ = 2}}{{\text{5}}^o}C{\text{ = 25 + 273 = 298K}}$
Using the formula for work done, substitute the given values in the formula.
\begin{array}{*{20}{l}} { \Rightarrow Work{\text{ }}done{\text{ }} = - nRT} \\\ = - 0.893 \times 8.314 \times 298 \\\ {\text{ = - 2212}}{\text{.47 joule }} \\\ \approx {\text{ - 2212}}{\text{.22 joule }} \\\ \end{array}
Thus, the $work\;done = -2212.22 J$
Final answer: Correct option: $D= -2212.22 \;J$ .

Additional Information:
To calculate the work done in calories, the value of the Universal gas constant (R) changes to $1.987 cal/Kmol$ . Rest all the values of the formula will be the same.

Note:
As the beaker was open, the pressure was 1atm and the change in volume was constant. The work done is negative so the work done is on the system. In this case energy is added to the system. If the work done is positive, the work is done by the system. In such cases the energy is spent by the system.