Question

A strip of magnesium of mass 15g is dropped into an open beaker of dilute hydrochloric acid. The work done by the system as a result of the reaction is :
[The atmospheric pressure is 1.0 atm and temperature is $25{}^\circ C$ ]
(A) - 1.548 KJ
(B) + 1.354 KJ
(C) + 1.348 KJ
(D) None of these

Answer 1

For work to occur, the force must have to be exerted on the body causing it to move. The work done is simply the product of force and displacement caused by the application of the force. Here the applied force is the pressure and displacement is the volume change. Therefore,
$W=- P\Delta V$

Complete Step by step answer:
- We will start the solution by writing the chemical reaction.
$Mg(s)+2HCl\to MgC{{l}_{2}}+{{H}_{2}}$
- According to the question, we have 15 g of magnesium. From this, we can calculate the number of moles of magnesium before reaction as-
$\text{Number of moles of magnesium}=\dfrac{\text{Given mass}}{\text{Molar mass}}=\dfrac{15}{24.325}moles$
- It is clear from the reaction that one mole of magnesium in reaction with two moles of hydrochloric acid gives one mole of hydrogen and one mole of magnesium chloride. As the reactants are in the solid and liquid state and the product formed is a solid salt, hence they will not make a significant volume change. Volume change is mainly given by the volume of hydrogen gas formed in the product as it is a gaseous state. Therefore,
$\Delta V={{(V)}_{{{H}_{2}}}}$
- After the reaction, the number of moles of magnesium left will be zero and the number of moles of hydrogen will be equal to the number of moles of magnesium before reaction, i.e
$\text{Number of moles of }{{\text{H}}_{2}}=\dfrac{15}{24.325} moles$
- From the mole concept, we know that,
1 mole of gas contains = 22.4 l
There. In $\dfrac{15}{24.325}$ moles of gas will have $=\dfrac{15}{24.325}\times 22.4=13.81l$
- Now calculating the work done,
$W=- P\Delta V=- (1atm)\times 13.81l=- 13.81\times 101.3J=- 1.548KJ$
Work done is taken negatively as the work is done by the system.
So the correct answer is option A.

Note: Alternatively, we can also use the ideal gas equation for the calculation of volume.
$PV=nRT$
The pressure given is 1 atm;
Number of moles $=\dfrac{15}{24.325}$
R = 8.314
T = 25 + 273 = 298 K
$\Rightarrow 1(V)=\dfrac{15}{25.352}\times 8.314\times 298=1548.8$
From the formula of work done and using the volume calculated in the earlier step, we can find the answer-
Work done = - PV
$=- 1\times 1548.88=- 1548J=- 1.548KJ$