Consider the following system at equilibrium where H° = 16.1 kJ, and Kc = 6.50×10-3, at 298 K. 2NOBr(g) 2NO(g) + Br2(g)If the VO

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Consider the following system at equilibrium where H° = 16.1 kJ, and Kc = 6.50×10-3, at 298 K. 2NOBr(g) 2NO(g) + Br2(g)If the VO

LUME of the equilibrium system is suddenly increased at constant temperature:The value of Kc:A. increases. B. decreases. C. remains the same.The value of Qc:A. is greater than Kc. B. is equal to Kc. C. is less than Kc.The reaction must:A. run in the forward direction to reestablish equilibrium. B. run in the reverse direction to reestablish equilibrium. C. remain the same. It is already at equilibrium.The number of moles of Br2 will:A. increase. B. decrease. C. remain the same.

Chemistry

1 answer:

Norma-Jean [14] · Answer 1 · 2022-03-09T18:47:38+03:00

Answer:

1) The value of Kc : (C.) remains the same

2) The value of Qc : (A.) is greater than Kc

3) The reaction must : (B.) run in the reverse direction to reestablish equilibrium.

4) The number of moles of Br2 will : (B.) decrease

Explanation:

Value of concentration equilibrium constant Kc depends only on temperature. Since temperature remains constant, therefore, Kc remains constant. Decrease in volume means increase in pressure. Increase in pressure favors the side with less gaseous species. Hence, increase in pressure will favor the reverse reaction towards reactants.