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Consider the following reaction at equilibrium. What effect will removing NO_2 have on the system? SO_2 (g) + NO_2 (g) = SO_3 (g) + NO (g) The reaction will shift in the detection of products. The reaction will shift to decrease the pressure. The reaction will shift in the direction of reactants. The equilibrium constant will decrease. No change will occur since SO_3 is not included in the equilibrium expression. Express the equilibrium constant for the following reaction CH_4 (g) + 2 O_2 (g) = CO_2 (g) + 2 H_2 O (g) K = [CO_2] [H_2 O]^2/[CH_4] [O_2]^2 K = [CO_2] [2H_2 O]^2/[CH_4] [2O_2]^2 K = [CH_4] [O_2]^2/[CO_2] [H_2 O] K = [CH_4] [2O_2]^2/[CO_2] [2H_2 O]^2 K = [CH_4] [O_2]/[CO_2] [H_2 O]
Consider the following reaction at equilibrium. What effect will removing NO_2 have on the system? SO_2 (g) + NO_2 (g) = SO_3 (g) + NO (g) The reaction will shift in the detection of products. The reaction will shift to decrease the pressure. The reaction will shift in the direction of reactants. The equilibrium constant will decrease. No change will occur since SO_3 is not included in the equilibrium expression. Express the equilibrium constant for the following reaction CH_4 (g) + 2 O_2 (g) = CO_2 (g) + 2 H_2 O (g) K = [CO_2] [H_2 O]^2/[CH_4] [O_2]^2 K = [CO_2] [2H_2 O]^2/[CH_4] [2O_2]^2 K = [CH_4] [O_2]^2/[CO_2] [H_2 O] K = [CH_4] [2O_2]^2/[CO_2] [2H_2 O]^2 K = [CH_4] [O_2]/[CO_2] [H_2 O]
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Jean KeelingLv2
20 Apr 2019
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