Consider this reaction. 2C(s)+ O_2(g) 2CO(g) What is the equilibrium expression
ID: 501931 • 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 according 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 container, 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) decreasing the temperature (D) decreasing the pressureExplanation / Answer
13. ANS: B, kc = [CO]2/[C]2[O2]
14.ANS:D, the chemical equilibrium in a reversible reaction is the state at which both forward and backward reactions or two reactions occur at same speed.
15.ANS: C, 1.7
kc for given recation = [CH3OH]/[CO][H2]2, from the given concentrations
kc= 0.06/(0.40)(0.30)2
= 1.6667= 1.7
16.ans:C , decreasing the temperature, given reaction is exothermic