1. The equilibrium constant of the balance reaction below is 1.6 x 10-2 at 25 oC and 2.7 x 10-5 at 125 oC.
AB (g) + C (g) 3D (g) Which of the following is wrong for this reaction?
a) Adding catalyst does not change the equilibrium constant of the reaction.
b) The reaction is exothermic.
c) If the total pressure is increased, the partial pressure of the product increases
d) Increasing the concentration of C gas in the environment increases the concentration of D gas.
e) Increasing the volume affects balance
2. Which of the following reduces the occurrence of HI in the given reaction?
H2 (g) + I2 (g) 2HI (g) ΔH = 52.96 kJ
a) Increasing the amount of H2 b) Removing HI from the environment c) Lowering the temperature
d) Halving the volume of the reaction vessel e) Increasing the amount of I2
1. a is correct.
b. You will need to work this out. The van't Hoff equation is
ln(k2/k2) = delta Ho(1/T1-1/T2)/R. Plug in the numbers and solve for delta Ho.
c is wrong. When P is increased the reaction will shift to the side with the smaller number of mols.
d is correct.
I don't know what you mean by answer e.
2. I assume you mean by "reduces the occurence of HI" causes the reaction to shift to the left. c is the correct answer but you should understand why this is correct and the others answers are not.
To determine which statement is incorrect for each question, let's go through the options one by one.
1. The equilibrium constant (K) of a reaction represents the ratio of the concentrations (or partial pressures) of products to reactants at a given temperature. It is a constant value that does not change with the addition of a catalyst (option a). Therefore, if adding a catalyst does not change the equilibrium constant, it is correct.
Exothermic reactions release heat energy and usually favor the formation of products at lower temperatures (option b). So, if the equilibrium constant is higher at higher temperature (2.7 x 10-5 at 125 oC) compared to the equilibrium constant at a lower temperature (1.6 x 10-2 at 25 oC), it implies that the reaction is indeed exothermic. Hence, option b is correct.
According to Le Chatelier's principle, if the total pressure is increased, the system will respond by shifting in the direction that reduces the pressure. Since the product side has 3 moles of gas (3D), and the reactant side only has 1 mole of gas (AB + C), increasing the total pressure will favor the production of more product (option c). Therefore, option c is correct.
By changing the concentration of a reactant, we can affect the equilibrium position. Increasing the concentration of C gas in the environment would shift the reaction to the right (product side) to restore equilibrium. Since D is a product of the reaction, its concentration would increase (option d). Thus, option d is correct.
Lastly, increasing the volume of a system affects the equilibrium position. If the volume is increased, the system tends to shift in the direction that increases the number of moles of gas to occupy the larger volume. In this case, the product side has more moles of gas (3 moles of D) compared to the reactant side (1 mole of AB + 1 mole of C). Therefore, increasing the volume would shift the reaction to the right (option e). Hence, option e is also correct.
Based on the explanations above, the answer to question 1 is option a because adding a catalyst does not change the equilibrium constant.
Now, let's move on to question 2.
2. The given reaction is an equilibrium between H2 (reactant) and HI (product), where the forward reaction is exothermic (ΔH = -52.96 kJ).
To reduce the occurrence of HI (option a), we need to favor the reverse reaction. According to Le Chatelier's principle, increasing the concentration of H2 will shift the equilibrium towards the reactant side, reducing the concentration of HI.
Removing HI from the environment (option b) would disturb the equilibrium, but it does not necessarily reduce the occurrence of HI. Hence, option b is not correct.
Lowering the temperature (option c) generally favors the exothermic direction of the reaction, which is the forward reaction in this case. Therefore, it would increase the occurrence of HI. Thus, option c is not correct.
Halving the volume of the reaction vessel (option d) will increase the pressure in the system, favoring the side with fewer moles of gas, which is the reactant side (H2 + I2). Hence, it would lead to a reduction in the occurrence of HI. Therefore, option d is correct.
Similarly, increasing the amount of I2 (option e) would shift the equilibrium towards the product side, favoring the formation of HI. Therefore, it would not reduce the occurrence of HI. Hence, option e is not correct.
Based on the explanations above, the incorrect statement for question 2 is option c: lowering the temperature reduces the occurrence of HI.