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Question
Calculate the amount of work done in the
1) Oxidation of 1 mole HCl(g) at 200 °C according to reaction.
4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g)
2) Decomposition of one mole of NO at 300 °C for the reaction
2NO(g) → N2(g) + O2(g)
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Solution
Given:
1) Oxidation of 1 mole HCl(g)
Temperature = T = 200 °C = 473 K
2) Decomposition of one mole of NO
Temperature = T = 300 °C = 573 K
To find: Work done
Formula: W = - ΔngRT
Calculation:
1) The given reaction is for 4 moles of HCl. For 1 mole of HCl, the reaction is given as follows:
HCl(g) + `1/4` O2(g) → `1/2` Cl2(g) + `1/2` H2O(g)
Now,
Δng = (moles of product gases) - (moles of reactant gases)
Δng = `(1/2 + 1/2) - (1 + 1/4) = - 0.25` mol
Hence,
W = - ΔngRT
= -(- 0.25 mol) × 8.314 J K-1 mol-1 × 473 K
= + 983 J
2) The given reaction is for 2 moles of NO. For 1 mole of NO, the reaction is given as follows:
NO(g) → `1/2` N2(g) + `1/2` O2(g)
Now,
Δng = (moles of product gases) - (moles of reactant gases)
Δng = = `(1/2 + 1/2) - 1 = 0` mol
Hence,
W = - ΔngRT
= - 0 mol × 8.314 J K-1 mol-1 × 573 K
= 0 kJ
No work is done (since W = 0).
∴ The work done is +983 J. The work is done on the system.
∴ The work done is 0 kJ. There is no work done.
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