Consider the following exothermic reaction: 2HI (g) H 2 (g) + I 2 (g) A) What wi
ID: 510957 • Letter: C
Question
Consider the following exothermic reaction:
2HI (g) H2 (g) + I2 (g)
A) What will happen to the reaction mixture at equilibrium if an inert gas is added?
[1] The equilibrium will shift to the left.
[2] The equilibrium will shift to the right.
[3] There is no effect on the equilibrium.
[4] The reaction will stop.
[5] None of the above.
B) What will happen to the reaction mixture at equilibrium if the volume of the container is increased?
[1] The equilibrium will shift to the left.
[2] The equilibrium will shift to the right.
[3] There is no effect on the equilibrium.
[4] The reaction will stop.
[5] None of the above.
C) What will happen to the reaction mixture at equilibrium if the pressure is increased?
[1] The equilibrium will shift to the left.
[2] The equilibrium will shift to the right.
[3] There is no effect on the equilibrium.
[4] The reaction will stop.
[5] None of the above.
D) What will happen to the reaction mixture at equilibrium if some H2 (g) is removed?
[1] The equilibrium will shift to the left.
[2] The equilibrium will shift to the right.
[3] There is no effect on the equilibrium.
[4] The reaction will stop.
[5] None of the above.
E) What will happen to the reaction mixture at equilibrium if the temperature is increased?
[1] The equilibrium will shift to the left.
[2] The equilibrium will shift to the right.
[3] There is no effect on the equilibrium.
[4] The reaction will stop.
[5] None of the above.
Can you please explain how to get to the answer?
Explanation / Answer
(A).
2HI (g) H2(g) + I2(g)
Answer is [3] There is no effect on the equilibrium.
Add an inert gas (one that is not involved in the reaction) to the constant-volume reaction mixture: This will increase the total pressure of the system, but will have no effect on the equilibrium condition. That is, there will be no effect on the concentrations or the partial pressures of reactants or products.
(B)
2HI (g) H2(g) + I2(g)
Answer is [3] There is no effect on the equilibrium.
Change the volume of the system: When the volume is changed, the concentrations and the partial pressures of both reactants and products are changed. If the volume is decreased, the reaction will shift towards the side of the reaction that has fewer gaseous particles. If the volume is increased, the reaction will shift towards the side of the reaction that has more gaseous particles.
In this case both side moles are same.
(C)
2HI (g) H2(g) + I2(g)
Answer is [3] There is no effect on the equilibrium.
When there are the same number of molecules on both sides, increasing the pressure has no effect on the position of the equilibrium. Because there are equal numbers of molecules on both sides, the equilibrium cannot move in any way that will reduce the pressure again.
(D).
2HI (g) H2(g) + I2(g)
Answer is [2] The equilibrium will shift to the right.
When H2(g) is removed, the concentration of H2(g) is decreased, according to Le Chatelier principle, the position of equilibrium will move so that the concentration of H2(g) increases again. More HI will react to replace the H2(g) that has been removed. The position of equilibrium moves to the right.
(E).
2HI (g) H2(g) + I2(g) (Exothermic reaction)
Answer is [1] The equilibrium will shift to the left.
If the temperature is increased, then the position of equilibrium will move so that the temperature is reduced again.
Increasing the temperature of a system in dynamic equilibrium favors the endothermic reaction. The system counteracts the change by absorbing the extra heat.
Forward reaction is exothermic and hence the backward reaction is endothermic.
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