Q. For a reaction with a rate constant of 0.5 s^-1, how long will it take for the concentration of a reactant to decrease to 25% of its initial value in a first-order reaction?
A.
1.386 seconds
B.
2 seconds
C.
4 seconds
D.
8 seconds
Solution
For a first-order reaction, the time to reach 25% of the initial concentration is t = (ln(1/0.25))/k = (ln(4))/0.5 = 2.772/0.5 = 5.544 seconds.
Q. For a reaction with an activation energy of 50 kJ/mol, what is the effect of a 10 kJ/mol increase in activation energy on the rate constant at a constant temperature?
A.
Rate constant increases
B.
Rate constant decreases
C.
Rate constant remains the same
D.
Rate constant becomes zero
Solution
An increase in activation energy decreases the rate constant according to the Arrhenius equation, k = Ae^(-Ea/RT), where an increase in Ea results in a smaller k.
Q. For a reaction with an activation energy of 50 kJ/mol, what is the effect of increasing the temperature on the rate constant?
A.
Rate constant decreases
B.
Rate constant increases
C.
Rate constant remains the same
D.
Rate constant becomes zero
Solution
According to the Arrhenius equation, an increase in temperature results in an increase in the rate constant, k, as it provides more energy to overcome the activation energy barrier.
Q. For a reaction with an activation energy of 50 kJ/mol, what is the effect of increasing the temperature from 300 K to 350 K on the rate constant?
A.
Rate constant decreases
B.
Rate constant remains the same
C.
Rate constant increases
D.
Rate constant doubles
Solution
According to the Arrhenius equation, an increase in temperature generally increases the rate constant due to the exponential dependence on temperature.
Q. For a redox reaction, if the standard reduction potentials are E°(A/B) = 0.80 V and E°(C/D) = 0.40 V, which species is the stronger oxidizing agent?
A.
A
B.
B
C.
C
D.
D
Solution
The stronger oxidizing agent is the one with the higher reduction potential. Here, A/B has a higher E° than C/D.
Q. For a zero-order reaction, if the initial concentration is 0.5 M and the rate constant is 0.1 M/s, how long will it take for the concentration to drop to 0.2 M?
A.
3 s
B.
5 s
C.
7 s
D.
10 s
Solution
For a zero-order reaction, [A] = [A]0 - kt. Thus, 0.2 M = 0.5 M - (0.1 M/s)t, leading to t = (0.5 - 0.2) / 0.1 = 3 s.
Q. For an ideal gas, which equation relates the change in internal energy to heat and work?
A.
ΔU = Q + W
B.
ΔU = Q - W
C.
ΔU = W - Q
D.
ΔU = Q * W
Solution
The first law of thermodynamics states that the change in internal energy (ΔU) is equal to the heat added to the system (Q) minus the work done by the system (W).
Q. For the equilibrium reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), what will happen if the volume of the container is increased?
A.
Shifts to the right
B.
Shifts to the left
C.
No effect
D.
Increases the reaction rate
Solution
Increasing the volume decreases the pressure, which shifts the equilibrium to the left, favoring the side with more moles of gas (3 moles of reactants).
Q. For the reaction 2NO(g) + O2(g) ⇌ 2NO2(g), if the initial concentrations are [NO] = 0.5 M and [O2] = 0.2 M, what is the equilibrium concentration of NO2 if Kc = 10?
A.
0.1 M
B.
0.2 M
C.
0.5 M
D.
0.4 M
Solution
Using the expression Kc = [NO2]^2 / ([NO]^2[O2]), we find that at equilibrium [NO2] = 0.4 M.
