Answer the following question. Calculate the work done in the decomposition of 132 g of NH4NO3 at 100 °C. NH4NO3(s) → N2O(g) + 2H2O(g) State whether work is done on the system or by the system. - Chemistry

Sum

Answer the following question.

Calculate the work done in the decomposition of 132 g of NH4NO3 at 100 °C.

NH4NO3(s) → N2O(g) + 2H2O(g)

State whether work is done on the system or by the system.

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:

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.

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 4.08 | Page 88
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