An adiabatic container fitted with a movable adiabatic piston (operating at 1 atm) is filled with 2 L of 2 M H2O2(aq) solution at 300 K. If H2O2 dissociates following first order decay with a half life of 10 min then change in internal energy in first 20 min will be [R = 2 Cal/mol K.]

Engineering

Chemistry

Integrated Rate Law

First Law and Various Process

Question

An adiabatic container fitted with a movable adiabatic piston (operating at 1 atm) is filled with 2 L of 2 M H₂O₂(aq) solution at 300 K. If H₂O₂ dissociates following first order decay with a half life of 10 min then change in internal energy in first 20 min will be [R = 2 Cal/mol K.]

1800 Cal

900 Cal

450 Cal

225 Cal

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Initial moles of H₂O₂ = 2 × 2 = 4
Moles of H₂O₂ left after 20 min. = 4 × $\frac{1}{2^{2}}$ = 1
(2 half lives)
Moles of H₂O₂ decomposed in 20 min. = 4 – 1 = 3
H₂O₂ → H₂O + $\frac{1}{2}$ O₂
Moles of O₂ formed during 20 min. = $\frac{3}{2}$
Volume change = $\frac{\frac{3}{2} \times R \times T}{P}$
q = 0 ΔU + W
W = – P_ext (V₂ – V₁) = – P_ext ΔV = $- \frac{3}{2} \times 2 \times 300$ = – 900 Cal.