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For the reaction C(s) +CO(2)(g) rarr 2CO...

For the reaction `C(s) +CO_(2)(g) rarr 2CO(g), k_(p)=63` atm at 100 K. If at equilibrium `p_(CO)=10p_(CO_(2)) ` then the total pressure of the gases at equilibrium is

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The correct Answer is:
6.93
`C(s) + CO_(2)(g) to 2CO(g)`
Appluying law of mass action,
`K_(P) = (P_(CO))^(2)/P_(CO_(2))` or `63 =(10P_(CO_(2))^(2))/P_(CO_(2))`
(Given `K_(p) = 63, P_(CO) = 10 P_(CO_(2))`)
`P_(CO) = 10 P_("total") = P_(CO_(2)) + P_(CO) = 0.63 + 6.3 = 6.93` atm
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