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For the reaction 2CO +O(2) rarr 2CO(2)...

For the reaction
`2CO +O_(2) rarr 2CO_(2), DeltaH =- 560 kJ`,
`2mol` of `CO` and `1mol` of `O_(2)` are taken in a container of volume `1L`. They completely form `2 mol` of `CO_(2)`. The gaseous deviate appreciably from ideal behaviour. If the pressure in the vessel changes from `70` to `40atm`, find the magnetic (absolute) value of `DeltaU at 500K. (1 L-atm = 0.1 kJ)`

Text Solution

Verified by Experts

`DeltaH=DeltaU+Delta(PV)=DeltaU+P_(2)V_(2)-P_(1)V_(1) ...(1)`
We cannot use `DeltaH=DeltaU+DeltanRT` as `Pvcancel(=)nRT` since gases deviates from ideal behaviour thus,
`Detla(PV)=P_(2)V_(2)-P_(1)V_(1)=40xx1-70xx1=-30litre-atm`
`=-3.0kJ`
`:. `By `eq.(1) -560=DeltaU-3.0`
`:. DeltaU=-557kJ`
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Knowledge Check

  • For the reaction, 2CO + O_(2) to 2CO_(2), DeltaH = -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 DeltaU at 500 K.

    A
    -557kJ
    B
    +557kJ
    C
    -5570J
    D
    +5570J

  • For the reaction, 2CO+O_(2)rarr2CO_(2), DeltaH=560 kJ . Two moles of CO and one mole of O_(2) are taken in a container of volume 1L. 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 DeltaU at 500 K (1 L atm = 0.1 kJ)

    A
    -557
    B
    575
    C
    585
    D
    595

  • For the reaction 2CO(g)+O_(2)(g)rarr 2CO_(2), DeltaH=-500kJ. Two moles of CO and one mole of O_(2) are taken in a container of volume 2 L. They completely from two moles of CO_(2) , the gas deviate appreciably from ideal behaviour. If pressure in vessel change from 35 to 20 atm. Find the magnitude of DeltaU" at "500K . (Assume 1 L - atom = 0.1 kJ)

    A
    503 kJ
    B
    400 kJ
    C
    480 kJ
    D
    320 kJ