Consider the following system at equilibrium where Delta H degree = -16.1 kJ, an
ID: 487589 • Letter: C
Question
Consider the following system at equilibrium where Delta H degree = -16.1 kJ, and K_c = 154, at 298 K. 2 NO(g) + Br_2(g) 2NOBr(g) If the VOLUME of the equilibrium system is suddenly decreased at constant temperature: The value of K_c A. increase B. decreases. C. remains the same. The value of Q_c A. is greater than K_c. B. is equal to K_c. C. is less than K_c. 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 Br_2 will: A. increase. B. decrease. C. remain the same.Explanation / Answer
a)
Kc --> it is always constant as far as Temperature is constant, so the value remainst he sam, choose C
b)
Qc ---> [products]/[reactants]
For this...
there are 3 moles in left and 2 in right..
Then, products increases, reactants decrease
Qc ---> [products]/[reactants]
Qc will INCREASE and be greater than Kc
c)
so.. a decrease in volume, will increase pressure, so this favours the LEAST mol formation, which is 2 moles, that is, the products, the reaction will favour in the FORWARD direction
d)
Moles of Br2, which is in the left, i.e. a reactant, will DECREASE in order to for mmore NOBr