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 VOLUME 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.

Respuesta :

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.

1) The value of Kc should remain  the same

2) The value of Qc should be greater than Kc

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

4) The number of moles of Br2 will decrease

Concentration equilibrium:

The value of concentration equilibrium should remain the same Kc. It is based only on the temperature. In the case when the temperature should be the same so the kc should be the same. In the case when there is a reduction in the volume so it rises the pressure. Due to increase in pressure it favored the slide having less gaseous species.

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