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 at817^(@)C and a total pressure of 5 atm ? (b) At what total pressure, the gas contains 5 % CO_(2) by voume ?

At 817^(@)C, K_(p) for the reaction between CO_(2) (g) and excess hot

1.	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 at817^(@)C and a total pressure of 5 atm ? (b) At what total pressure, the gas contains 5 % CO_(2) by voume ?
Answer» Solution :(a) ` CO_(2) + C(g) hArr2 CO (g)` Supposeat equilibrium , pressure of `CO_(p_CO_(2)) = p "atm "` Then pressureof ` 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 = (366)/(0 0821 xx ( 817 + 273)) = 0041 " mol"L ^(-1)` Molar conc. of ` CO_(2) = (134)/(00821 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 also5 %of the total pressure . Thus, if P is the total pressure , then at equilibrium , ` p_(CO_(2)) =005 P and p_(CO) = 0 95 P ` ` K_(P)= (p_(CO)^(2))/p_(CO^(2))= (095 P)^(2)/((0 05 P)) = 10 or 18* 05 P= 10 or P = 0* 554 "atm "`