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Grade upto college level Physical Chemistry

For the reaction, 2CO + o2 → 2CO2; ∆H = -560kJ. Two moles of CO and one mole of 0, are taken in a container of volume 1 L. They completely form two moles of CO2, 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 ∆U at 500K. (1 L atm = 0.l kJ)

Profile image of Shane Macguire
12 Years agoGrade upto college level
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1 Answer

Profile image of Deepak Patra
12 Years ago
Hello Student,
Please find the answer to your question
∆H = ∆U + ∆(PV) = ∆U + V ∆P (∵ ∆V = 0)
or ∆U = ∆H – V ∆P = - 560 – [1(40 - 70) * 0.1]
= - 560 + 3 = - 557 kJ mol-1
So the magnitude is 557 kJ mol-1.
∵ ∆G° = - 2.303RT log Kp at equilibrium ∆G° = 0
∴ - 2.303RT log Kp = 0
Log Kp = 0 or Kp = 1

Thanks
Deepak patra
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