The following reaction is at equilibrium: N_2(g) +3H_2(g) doubleheadarrow 2NH_3(
ID: 513777 • Letter: T
Question
The following reaction is at equilibrium: N_2(g) +3H_2(g) doubleheadarrow 2NH_3(g). If we increase the volume while holding the temperature constant, in what direction will the reaction move to reestablish equilibrium? (towards reactants or products?) For the reaction: 2NO(g) + O_2(g) doubleheadarrow 2NO-2(g), which accurately reflects the changes in concentration that will occur if O_2 is added to disturb the equilibrium? The following reaction is at equilibrium: N_2(g) + O_2(g) doubleheadarrow 2 NO(g). delta H' = +180.8 kJ. In what direction will the reaction move to reestablish equilibrium if the temperature is decreased? (reactants, products, or no change) For the reaction: A_2(g) + B_2(g) doubleheadarrow 2AB(g), the concentration-time graph is as follows: What is the value of the equilibrium constant?Explanation / Answer
1. Since the number of moles in the products side is less than that of the reactants in the reaction given, the volume of products is greater than the reactants. According to the Le Chatelier's principle, on increasing the volume, the reaction moves towards the side which has lower volume. In our case, the products have lower volume. Hence, the reaction will move towards the right side.
2. In case of increasing the concentration of Oxygen, the reaction will move in the forward direction as the concentration of Oxygen is lower on the products side. Hence, the concentration of NO2 would increase but the concentration of NO would decrease. Hence, the choice E.
On the other hand, if we decrease the concentration of Oxygen then the reaction will move in the reverse direction so as to increase the concentration of NO and decreasing the concentration of NO2.
3. In case of endothermic reaction as can be seen due to the positive sign of the enthalpy of reaction, on increasing the temperature the reaction in the direction of decreasing heat. In endothermic reactions, the products side have lower heat. Hence, the reaction will move in the forward direction.
4. Keq = [AB]2/[A]*[B]
From the graph, the concentration of AB at equilibrium is equal to 1. While the value of [A]*[B] is equal to 0.5. Substituting these values in the equation, we get Keq= 1/0.5 = 2
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