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) - Chemistry

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Sum

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.

Concept: Enthalpy (H)
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Chapter 4: Chemical Thermodynamics - Exercises [Page 88]

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Balbharati Chemistry 12th Standard HSC for Maharashtra State Board
Chapter 4 Chemical Thermodynamics
Exercises | Q 4.14 | Page 88

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