The synthesis of CH_3OH(g) from CO(g) and H_2(g) is represented by the equation

Question

The synthesis of CH_3OH(g) from CO(g) and H_2(g) is represented by the equation above. The value of K_c for the reaction at 483 K is 14.5 Which of the following explains the effect on the equilibrium constant, K_c, when the temperature of the reaction system is increased to 650 K? (A) K_c will increase because the activation energy of the forward reaction increases more than that of the reverse reaction. (B) K_c will increase because there are more reactant molecules than product molecules (C) K_c will decrease because the reaction is exothermic (D) K_c is constant and will not change. A 1.0 mol sample of CO(g) and a 1.0 mol sample of H_2(g) are pumped into a rigid, previously evacuated 2.0L reaction vessel at 483 K. Which of the following is true at equilibrium? (A) [H_] = 2[CO] (B) [H_2]

Explanation / Answer

21)
Forward reaction is exothermic in nature
we are increasing temperature or adding heat here
so, according to Le Chatellier's principle,
equilibrium will move in direction which absorbs heat
hence, backward reaction will be favoured
Equilibrium moves to reactant side
Kc will decrease
Answer: C

22)
initially,
[CO] = [H2]

but to reach equilibrium [H2] used = 2*[CO] used

so, at equilibrium
[CO] > [H2]
Answer: B

23)
reducing volume to half will double the pressure
Answer: C

24)
since reaction is exothermic, more energy is released than abdorbed
breaking bond requires energy
Answer: C

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