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Calculate the work done in the decomposition of 132 g of NH⁢4⁢NO3 at 100°C. NH⁢4⁢NO⁢3⁢ (s) -> N⁢2⁢O⁢ (g) + 2⁢H⁡2⁢O⁢ (g) State whether work is done on or by the system. - Chemistry

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Question

Calculate the work done in the decomposition of 132 g of \[\ce{NH4NO3}\] at 100°C.

\[\ce{NH4NO3_{(s)} -> N2O_{(g)} + 2H2O_{(g)}}\]

State whether work is done on or by the system.

Numerical
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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:

Molar mass of NH4NO3 = (2 × 14) + (3 × 16) + (4 × 1) = 80 g mol−1

Moles of NH4NO3 = n = `(132 "g")/(80 "g mol"^-1)` = 1.65 mol

The given reaction is for 1 mole of NH4NO3. For 1.65 moles of NH4NO3, the reaction is given as follows:

\[\ce{1.65 NH4NO3_{(s)} -> 1.65 N2O_{(g)} + 3.30 H2O_{(g)}}\] 

Now, 

Δng = (moles of product gases) − (moles of reactant gases)

Δng = (1.65 + 3.30) − 0 = +4.95 mol    ...(∵ NH4NO3 is in solid state)

Hence,

W = −Δng RT

= − (+ 4.95 mol) × 8.314 J K−1 mol−1 × 373 K

= −15350 J

= −15.35 kJ

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

The work done is −15.35 kJ. The work is done by the system.

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

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Balbharati Chemistry [English] Standard 12 Maharashtra State Board
Chapter 4 Chemical Thermodynamics
Exercises | Q 4.08 | Page 88

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