For the reaction, 2 CO + O 2 arrow 2 CO 2 ; Δ H =-560 kJ. Two moles of CO and one mole of O2 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 )

Question

For the reaction, 2 CO + O 2 arrow 2 CO 2 ; Δ H =-560 kJ. Two moles of CO and one mole of O2 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 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

From the definition of enthalpy, we have Δ H =Δ U +Δ( pV ) Δ H =Δ U + V Δ p + p Δ V ( at constant volume Δ V =0) Δ U =Δ H - V Δ p =-560+1 × 30 × 0.1( Dp =40-70=30 atm ) =-557 kJ

From the definition of enthalpy, we have ΔH=ΔU+Δ(pV) ΔH=ΔU+VΔp+pΔV( at constant volume ΔV=0) ΔU=ΔH−VΔp =−560+1×30×0.1(Dp=40−70=30atm) =−557kJ

From the definition of enthalpy, we have
$Δ H = Δ U + Δ (p V)$
$Δ H = Δ U + V Δ p + p Δ V ($ at constant volume $Δ V = 0)$
$Δ U = Δ H - V Δ p$
$= - 560 + 1 \times 30 \times 0.1 (D p = 40 - 70 = 30 a t m)$
$= - 557 k J$