Question

At `817^(@)C, K_(p)` for the reaction between `CO_(2) (g)` and excess hot graphite (s) is 10 atm
(a) What are the equilibrium concentrations of the gases at ` 817^(@)C` and a total pressure of 5 atm ?
(b) At what total pressure, the gas contains 5 % `CO_(2)` by voume ?

Answer 1

(a) ` CO_(2) + C(g) hArr 2 CO (g) `
Suppose at equilibrium , pressure of `CO_(p_CO_(2)) = p "atm "`
Then pressure of ` CO_(2) (p_(CO_(2)))= 5 - p " atm " `
` K_(p) (p_(CO)^(2))/(p_(CO_(2))) = (p^(2))/((5 - p))" " = 10 or p^(2) + 10 p - 50 = 0 `
or ` p = ( -b pm sqrt (b^(2) - 4ac))/(2a) = (-10 pm sqrt ( 100- (-200)))/2 = 3* 66" atm " `
Thus , at eqm. ` p _(CO) = 3 * 6 "atm " `
` p_(CO_(2)) = 5 - 3* 66 = 1* 34 "atm "`
Applying ` PV = nRT or n/V = P /(RT) , i.e ., "molar conc." = P/(RT) `
Molar conc. of `CO = (3*66)/(0* 0821 xx ( 817 + 273)) = 0*041 " mol"L ^(-1)`
Molar conc. of ` CO_(2) = (1*34)/(0*0821 xx 1090 ) = 0* 015 " mol" L^(-1)`
(b) When the gas contains 5 % `CO_(2)` by volume , this means that pressure exerted by `CO_(2) ` is also 5 % of the total pressure . Thus, if P is the total pressure , then at equilibrium , ` p_(CO_(2)) = 0*05 P and p_(CO) = 0* 95 P `
` K_(P) = (p_(CO)^(2))/p_(CO^(2))= (0*95 P)^(2)/((0* 05 P)) = 10 or 18* 05 P= 10 or P = 0* 554 "atm "`