Consider this reaction. 2C(s) + O_2(g) 2CO(g) What is the equilibrium expression
ID: 501928 • Letter: C
Question
Consider this reaction. 2C(s) + O_2(g) 2CO(g) What is the equilibrium expression for this reaction? (A) K_c = [CO]/[C][O_2] (B) K_c = [CO]^2/[C]^2[O_2] (C) K_c = [2CO]/[2C][O_2] (D) K_c = [CO]^2/[O_2] Chemical equilibrium is the result of (A) formation of products equal in mass to the mass of the reactants. (B) the unavailability of one of reactants. (C) a stoppage of further reaction. (D) opposing reactions attaining equal speeds. Carbon monoxide gas reacts with hydrogen gas at elevated temperatures to form methanol ac to this equation. CO(g) + 2H_2(g) CH_3OH(g) When 0.40 mol of CO and 0.30 mol of H_2 are allowed to reach equilibrium in a 1.0 L contain 0.060 mol of CH_3OH are formed. What is the value of K_c? (A) 0.50 (B) 0.98 (C) 1.7 (D) 5.4 Consider this reaction at equilibrium. 2SO_2(g) + O_2(g) 2SO_3(g) Delta H = -198 kJ Which of these changes would cause an increase in the SO_3/SO_2 mole ratio? (A) adding a catalyst (B) removing O_2(g) (C) decreasExplanation / Answer
13) Ans: Option (D)
14) Ans:option (A)
15) Ans: option (C)
16) Ans:option (C)