Question
Question: Calculate the work done when 11.29 g of Iron dissolves in HCl in i) a closed vessel ii) an open vess...
Calculate the work done when 11.29 g of Iron dissolves in HCl in i) a closed vessel ii) an open vessel at 25∘C. (Atomic mass of Fe = 56gmol−1)
Solution
First write the balanced reaction of iron with hydrochloric acid. For a closed vessel pressure will be constant. Then we can find the work done by following the formula. W=−PextΔV
Complete step by step solution:
We will calculate for the given reaction.
- The reaction between iron and hydrochloric acid gives ferrous chloride and hydrogen gas. The reaction can be given as under.
Fe+2HCl→FeCl2+H2
From the reaction, we can say that one mole of iron produces one mole of hydrogen gas.
- We know that 56 g of Fe is equivalent to 1 mole. So, 11.29 g of Fe will be equivalent to 5611.29=0.20 moles.
So, 0.20 moles of hydrogen gas will be produced when 11.29 g of Fe will react with HCl.
i) If the reaction is carried out in a closed vessel, then the change in volume will be zero.
Now, we know that work can be given as
W=−PextΔV
We know that ΔV = 0.
So, W = 0.
ii) If a reaction is carried out in an open vessel, then we need to calculate the change in volume.
- The newly formed hydrogen gas would occupy some volume. The volume occupied can be given as
V=PnRT
We know that number of moles n = 0.2 moles
Universal gas constant R = 0.0831 L atm/mol K
Temperature T = 298 K
and pressure P = 1 atm
So, we can write the above equation as
V=10.2×8.314×298=4.89L
Thus, we can say that the change in volume ΔV = 4.89 L
So, work W=−Pext×ΔV
So,
W=−1×4.89
So, W = -4.89 L atm
But we know that 1 L atm = 101.3 J.
So, -4.89 L atm = -4.89× 101.3 = -495.4 J
Thus, we obtained that the work done in a closed vessel will be zero and that in an open vessel will be equal to -495.4 J.
Note: Note that iron gets oxidized to +2 oxidation state and forms FeCl2 by the reaction with HCl. So, do not assume that iron gets oxidized to +3 oxidation state as it is also often seen in case of iron metal.