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Calculate the work done and comment on whether work is done on or by the system for the decomposition of 2 moles of NH4NO3 at 100 °CNH4NO3(s) → N2O(g) + 2H2O(g)

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Question

Calculate the work done and comment on whether work is done on or by the system for the decomposition of 2 moles of NH4NO3 at 100 °C
NH4NO3(s) → N2O(g) + 2H2O(g)

Sum
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Solution

Given: 

Decomposition of 1 mole of NH4NO3 

Temperature = T = 100 °C = 373 K

To find:

Work done and to determine whether work is done on the system or by the system

Formula: W = - Δ ngRT

Calculation:

The given reaction is for 1 mole of NH4NO3. For 2 moles of NH4NO3, the reaction is given as follows:

2NH4NO3(s) → 2N2O(g) + 4H2O(g)

Now, 
Δ n= (moles of product gases) - (moles of reactant gases)

Δ n= 6 - 0 = + 6 mol    (∵ NH4NO3 is in solid state)

Hence,

W = - Δ ngRT

= - (+ 6 mol) × 8.314 J K-1 mol-1 × 373 K

= - 18606.75 J

= - 18.61 kJ

Work is done by the system (since W < 0).

The work done is - 18.61 kJ. The work is done by the system.

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Chapter 4: Chemical Thermodynamics - Exercises [Page 89]

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Balbharati Chemistry [English] Standard 12 Maharashtra State Board
Chapter 4 Chemical Thermodynamics
Exercises | Q 4.2 | Page 89

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