1. For the exothermic reaction below, increasing the pressure would
N2(g)+3H2(g)⇄2NH3(g)
a. increase [H2]
b. increase [NH3]
c. increase [N2]*** Would it move to the side with fewer gas molecules?
d. have no effect
2. If K=100, then the value of K for the reverse reaction is
a. the same value
b. can only be determined by experimentation ****
c. the negative of the value for the forward reaction
d. 0.01
1. Yes, the reaction will move to the side with the fewer gas molecules but that isn't answer c. You have 4 mols gas on the left and 2 mols on the right so it will move to the right. The correct answer is b.
2. b is not correct. For the reverse reaction K is 1/100 = ?
2. is d
right. K rev = (1/Kfwd)
1. For the exothermic reaction N2(g) + 3H2(g) ⇄ 2NH3(g), increasing the pressure would have what effect on the concentrations of the reactants and products?
To determine the effect of increasing the pressure on the concentrations of the reactants and products, we need to consider Le Chatelier's Principle. According to this principle, if a stress is applied to a system at equilibrium, the system will shift in a way that alleviates the stress.
In this case, increasing the pressure would be considered a stress on the system. When the pressure is increased, the system will try to reduce the pressure by favoring the side of the reaction with fewer gas molecules.
Looking at the balanced equation, we see that there are four gas molecules on the left side (N2 and H2) and two gas molecules on the right side (NH3). Therefore, increasing the pressure would cause the system to shift to the side with fewer gas molecules, which is the left side.
So, increasing the pressure in this exothermic reaction would increase the concentration of N2, as indicated by option c in the question.
2. If K=100, what is the value of K for the reverse reaction?
The value of the equilibrium constant, K, is determined by the ratio of the concentrations or partial pressures of the products to the concentrations or partial pressures of the reactants at equilibrium. It does not depend on the specific direction of the reaction.
In this case, the given equilibrium constant, K, is 100 for the forward reaction. The reverse reaction is simply the reaction written in the opposite direction, so the equilibrium constant for the reverse reaction, denoted as K', would also be equal to 100.
Therefore, the answer to this question is option a: the value of K for the reverse reaction is the same as the value for the forward reaction.