Question

02 INTRO UNIV CHEM Il (LEC D1 Wi18) Te left Use the following equilibrium constants co2(g) + H2(g) <→ CO(g) + H2O(g) K,(400 K) 6.5 x 10 E(s) + 1 /202(g) <==> CO(g) K,(400 K) 1.3 x 1019 CO(g)+ 1/2 O2(g)CO2(g) K,(400 K) 2.8 x 102 CO(g) + 2 H2(g)CH3OH(g) K,(400 k) -2.3 H2(g) + 1 /202(g)-> H2O(g) (400 K)-1.8 x 1029 to determine the equilibrium constant for the following reaction. (You will not need to use all of the above equilibria.) 2 CO(g) + 2 H2(g)-12 02(g) <==> CO2(g) + CH3OH(g) (400 K) =

Answer 1

Target equation is 2CO_(g) + 2H_2(g) + 1/2 O_2 (g) CO_2 (g) + CH_3OH(g) k_P = ? K_P = if we analyse the K_p expression we need, co_2 and CH_3OH as product and CO, H_2 and O_2 as reagents. Select equation 3 and 4 from the given equation from eqn 3 CO_(g) + 1/2O_2(g) propositionallogic CO_(g) K_p = 2.8 times 10^32 K_p = P(CO_2(g))/P(co_(g)) mid dot P^1/2(O_2(g)) = 2.8 times 10^32 Eqn 4 CO_(g) + 2H_2(g) propositionallogic CH_3OH_(g) K_p = 2.3 k_P4 = P(CH_3OH_(g))/P_CO_(g) middot P_H_2(g) = 2.3 multiply K_P3 and K_p4 P(CO_2(g))/P(co_(g)) mid dot P^1/2(O_2(g)) middot P(CH_3OH_(g))/P_CO_(g) middot P_H_2(g) = 2.8 times 10^32 times 2.3 P(CO_2(g))/P(co_(g)) middot P(CH_3OH_(g))/P^2_(co_(g) middot P^2_(H_2(g)) middot P(O_2(g))^1/2 = K_p for the target equation therefore K_p = 2.8 times 10^32 times 2.3 K_p = 6.44 times 10^32