Question

Calculate the work done in the
decomposition of 132 g of NH4
NO3 at 100 °C.
NH4NO3 (s) : N2O(g) + 2 H2O(g)
State whether work is done on the system
or by the system.

Answer 1

Answer:

–15.35 kJ.

Explanation:

Given:

• Decomposition of 1 mole of NHąNO3
• 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 = -A ngRT

Calculation:

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

• Moles of NH4NO3 = n = 132g/80g 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.