Consider the following system at equilibrium where H: 182 kJ, and Kc = 7.00×10-5
ID: 541166 • Letter: C
Question
Consider the following system at equilibrium where H: 182 kJ, and Kc = 7.00×10-5, at 673 K. NH41(s)NH3(g HI(g) When some moles of NH41(s) are added to the equilibrium system at constant temperature: the value of KeA. increases. B. decreases C. remains the same. the value of Qe 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 restablish equilibrium. B. run in the reverse direction to restablish equilibrium. C. remain the same. It is already at equilibrium. the concentration of NH3 will:A. increase B. decrease. C. remain the same. Submit Answer Retry Entire Group 2more group attempts remainingExplanation / Answer
1. Kc value increases, a/c to Le Chatlier's principle addition of reactant shifts the direction to right side to decrease the amount of added reactant .
Value of Q is less than Kc in the case of forward reaction .
Equilibrium will shift in forward direction and concentration of NH3 increases
2 . Kc value increased due to the removal (concentration decreased) of NO, reaction moves in a direction to increase the concentration of NO, ie. , forward reaction, hence Kc increases
Qc is less than the Kc
Reaction shifts to right side and concentration of Br2 increases
3 . Kc remains same , since Kc is independent of volume
Qc is less than Kc since reaction moves in forward direction because if increase in volume shifts the the reaction towards more no of moles of gas.
And no of moles of Cl2 increases
4. Kc remains same and Qc is greater than the Kc since reaction will moves back (due to more no of gas molecules are present on reactant side) because of increase in volume .
Reaction moves in reverse direction and amount of H2 will increases.