Solutions

Solutions

Question

A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is

  • 1.8 atm

  • 3 atm

  • 0.3 atm

  • 0.3 atm

Answer

A.

1.8 atm

Kp depends upon the partial pressure of reactants and products so first calculate their partial pressure and then, calculate Kp,
stack space space space space space space space space space space space space space space space space CO subscript 2 with space Initial space space space space space space space space space space space 0.5 space atm below space left parenthesis straight g right parenthesis space space plus space straight C space left parenthesis straight s right parenthesis space rightwards harpoon over leftwards harpoon with space on top stack space 2 space CO with minus below left parenthesis straight g right parenthesis

At space equilibrium
space space space space space space space space space space space space left parenthesis 0.5 minus straight p right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space 2 space straight p space atm
This space is space straight a space case space of space hetergeneous space equilibrium.
straight C left parenthesis straight s right parenthesis space being space solid space is space not space considered
Total space pressure space of space CO subscript 2 space and space CO space gases.
straight p subscript CO subscript 2 space plus straight p subscript CO space equals space straight p subscript total end subscript
0.5 minus straight p space plus space 2 straight p space equals space 0.8 comma
straight p space equals space 0.3 space atm
therefore space straight p subscript CO subscript 2 end subscript space equals space 0.5 minus 0.3 space equals space 0.2 space atm
straight p subscript CO space equals space 2 straight p space equals space 0.6 space atm
straight K subscript straight p space equals space straight p subscript CO over straight p subscript CO subscript 2 end subscript space equals space fraction numerator 0.6 space straight x space 0.6 over denominator 0.2 end fraction space equals space 1.8 space atm

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