# 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 ga

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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 ?

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(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 "