For the reaction, $2 \mathrm{CO}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{CO}_{2} ; \Delta \mathrm{H}=-560 \mathrm{~kJ}$. Two moles of CO and one mole of $\mathrm{O}_{2}$ are taken in a container of volume 1 L . They completely form two moles of $\mathrm{CO}_{2}$, the gases deviate appreciably from ideal behaviour. If the pressure in the vessel changes from 70 to 40 atm, find the magnitude (absolute value) of $\Delta \mathrm{U}$ at 500 K . $(1 \mathrm{~L} \mathrm{~atm}=0.1 \mathrm{~kJ})$

JEE Advanced · 2006 · 09 Apr 2006 · Chemical Thermodynamics

For the reaction,

2 CO + O_{2} ⟶ 2 CO_{2}; Δ H = - 560 kJ

. Two moles of CO and one mole of

O_{2}

are taken in a container of volume 1 L . They completely form two moles of

CO_{2}

, the gases deviate appreciably from ideal behaviour. If the pressure in the vessel changes from 70 to 40 atm, find the magnitude (absolute value) of

Δ U

at 500 K .

(1 L atm = 0.1 kJ)

Answer: 557

Solution

Using the relationship between enthalpy and internal energy:

Δ H = Δ U + Δ (PV)

Since the gases deviate from ideal behavior, we use the actual pressure change:

Δ H = Δ U + V Δ P

- 560 = Δ U + 1 \times (40 - 70) \times 0.1

- 560 = Δ U + 1 \times (- 30) \times 0.1

- 560 = Δ U - 3

Δ U = - 557 kJ

The magnitude (absolute value) is 557 kJ.

JEE Advanced · 2006 · 09 Apr 2006 · Numerical answer

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