Consider the following system at equilibrium where H=-10.4 kJ, and Kc-55.6, at 6
ID: 1089249 • Letter: C
Question
Consider the following system at equilibrium where H=-10.4 kJ, and Kc-55.6, at 698 K. H2(g) + 12(g) 2HI(g) If the VOLUME of the equilibrium system is suddenly increased at constant t The value of Ke A. increases. emperature: B. decreases. C. remains the same. The value of QcA. is greater than Kc. B. is equal to Kc. C. is less than K 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 reaction must: The number of moles of I2 will: A. increase. B. decrease. C. remain the same.Explanation / Answer
H2(g) + I2(g) <======> 2HI(g)
Number of gaseous reactants = 2
Number of gaseous products = 2
The number of moles of gaseous reactants and products are same. Hence, volume change doesn't effect equilibrium.
The value of Kc : C. Remains same
The value of Qc: B is equal to Kc
The reaction must: C. Remains the aams . It is already at equilibrium.
The number of moles of I2 will: C remains the same.