Question bank/JEE Main/Chemistry

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:
dqN CO2, CO esa ifjofrZr gks tkrh gSA ij dqy nkc 0.8 atm gks
rks K

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CO2(g) + C(s)[IMAGE] 2CO (g)
0.5 atm
0.5-p 2p
Total pressure ¼dqy nkc½ = 0.5 - P + 2P = 0.8
P = 0.3
$K_{p} = \frac{P_{CO}^{2}}{P_{{CO}_{2}}} = \frac{(2P)^{2}}{(0.5 - P)} = \frac{(0.6)^{2}}{(0.5 - 0.3)}$
Kp = 1.8

Source: Grass-G / CH-26 Chemistry Paper 1 15 Jan.docx