Calculate the work done when 11.29 g of Iron dissolves in HCl in i) a closed vessel ii) an open vessel at ${25^ \circ }C$. (Atomic mass of Fe = 56$gmo{l^{ - 1}}$)
Answer
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Hint: 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. \[W = - {P_{ext}}\Delta V\]
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.
\[Fe + 2HCl \to FeC{l_2} + {H_2}\]
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 $\dfrac{{11.29}}{{56}} = 0.20$ 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
\[W = - {P_{ext}}\Delta V\]
We know that $\Delta V$ = 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
\[V = \dfrac{{nRT}}{P}\]
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
\[V = \dfrac{{0.2 \times 8.314 \times 298}}{1} = 4.89L\]
Thus, we can say that the change in volume $\Delta V$ = 4.89 L
So, work $W = - {P_{ext}} \times \Delta V$
So,
\[W = - 1 \times 4.89\]
So, W = -4.89 L atm
But we know that 1 L atm = 101.3 J.
So, -4.89 L atm = -4.89$ \times $ 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 $FeC{l_2}$ 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.
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.
\[Fe + 2HCl \to FeC{l_2} + {H_2}\]
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 $\dfrac{{11.29}}{{56}} = 0.20$ 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
\[W = - {P_{ext}}\Delta V\]
We know that $\Delta V$ = 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
\[V = \dfrac{{nRT}}{P}\]
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
\[V = \dfrac{{0.2 \times 8.314 \times 298}}{1} = 4.89L\]
Thus, we can say that the change in volume $\Delta V$ = 4.89 L
So, work $W = - {P_{ext}} \times \Delta V$
So,
\[W = - 1 \times 4.89\]
So, W = -4.89 L atm
But we know that 1 L atm = 101.3 J.
So, -4.89 L atm = -4.89$ \times $ 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 $FeC{l_2}$ 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.
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