Question

the half-life ($t_{1/2}$) of the catalyzed isomerization of cis-2-butene gas to produce trans-2-butene gas, represented above, was measured under various conditions, as shown in the table below.

trial number	initial $p_{cis-2-butene}$ (torr)	v (l)	t (k)	$t_{1/2}$ (s)
2	600.	2.00	350.	100.
3	300.	4.00	350.	100.
4	300.	2.00	365	50.

a. the reaction is first order. explain how the data in the table are consistent with a first-order reaction.

b. calculate the rate constant, $k$, for the reaction at 350. k. include appropriate units with your answer.

c. is the initial rate of the reaction in trial 1 greater than, less than, or equal to the initial rate in trial 2? justify your answer.

d. the half-life of the reaction in trial 4 is less than the half-life in trial 1. explain why, in terms of activation energy

Explanation:

Part a

Step1: Recall first-order half-life property

For a first - order reaction, the half - life formula is \(t_{1/2}=\frac{\ln2}{k}\), which means that the half - life of a first - order reaction is independent of the initial concentration (or in the case of gases, initial pressure) and also independent of the volume (since concentration is related to pressure and volume by \(PV = nRT\) or \(C=\frac{P}{RT}\)).

Step2: Analyze the table data

• In trials 1 and 2, the initial pressure (\(P_{cis - 2 - butene}\)) changes from 300 torr to 600 torr (doubles), but the volume (\(V = 2.00\ L\)) and temperature (\(T = 350\ K\)) are constant, and the half - life (\(t_{1/2}\)) remains 100 s.
• In trials 1 and 3, the volume changes from 2.00 L to 4.00 L (doubles), the initial pressure is constant (300 torr), and the temperature is constant (350 K), and the half - life remains 100 s.
• This shows that the half - life does not depend on the initial pressure (or concentration, since for a gas at constant T and V, \(P\propto n\) and \(C=\frac{n}{V}\)) or the volume (which would change the concentration if pressure was constant). So the data is consistent with a first - order reaction because the half - life is constant regardless of changes in initial pressure (trials 1 vs 2) and volume (trials 1 vs 3) at a constant temperature.

Part b

Step1: Recall first - order rate constant formula

For a first - order reaction, the relationship between half - life (\(t_{1/2}\)) and rate constant (\(k\)) is \(t_{1/2}=\frac{\ln2}{k}\), so we can solve for \(k\) as \(k = \frac{\ln2}{t_{1/2}}\).

Step2: Substitute values

At \(T = 350\ K\), from the table, \(t_{1/2}=100\ s\). \(\ln2\approx0.693\). So \(k=\frac{0.693}{100\ s}=6.93\times 10^{- 3}\ s^{-1}\).

Part c

Step1: Recall first - order rate law

The rate law for a first - order reaction is \(rate = k[cis - 2 - butene]\) (or in terms of pressure, since for a gas \(P\propto n\) and \(C=\frac{n}{V}\), \(rate=k'P_{cis - 2 - butene}\) where \(k'\) is a rate constant related to \(k\) by the ideal gas law).

Step2: Compare initial pressures

In trial 1, initial pressure \(P_1 = 300\) torr, in trial 2, initial pressure \(P_2=600\) torr. Since the rate is proportional to the initial pressure (for a first - order reaction with respect to \(cis - 2 - butene\)), and \(P_2 > P_1\), the initial rate in trial 2 is greater than the initial rate in trial 1. Mathematically, \(rate_1=kP_1\) and \(rate_2 = kP_2\), since \(P_2 = 2P_1\), \(rate_2=2rate_1\). So the initial rate in trial 1 is less than the initial rate in trial 2.

Part d

Answer:

Step1: Recall Arrhenius equation

The Arrhenius equation is \(k = A e^{-\frac{E_a}{RT}}\), where \(k\) is the rate constant, \(A\) is the pre - exponential factor, \(E_a\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature.

Step2: Relate rate constant to half - life

For a first - order reaction, \(t_{1/2}=\frac{\ln2}{k}\). So if \(t_{1/2}\) decreases, \(k\) increases.

Step3: Analyze temperature change

Trial 4 has a higher temperature (\(T = 365\ K\)) than trial 1 (\(T = 350\ K\)). According to the Arrhenius equation, as temperature \(T\) increases, the exponent \(-\frac{E_a}{RT}\) becomes less negative (since \(T\) is in the denominator), so \(e^{-\frac{E_a}{RT}}\) increases, and thus \(k\) increases. Since \(t_{1/2}\) is inversely proportional to \(k\) for a first - order reaction, an increase in \(k\) (due to an increase in temperature) leads to a decrease in \(t_{1/2}\). The activation energy \(E_a\) is a constant for a reaction. The higher temperature in trial 4 provides more molecules with energy greater than or equal to the activation energy, increasing the rate constant \(k\), and since \(t_{1/2}=\frac{\ln2}{k}\), a larger \(k\) leads to a smaller \(t_{1/2}\).

Part a Answer: The data is consistent with a first - order reaction because the half - life is constant (100 s) when initial pressure (trials 1 vs 2) and volume (trials 1 vs 3) change at constant T, which is a property of first - order reactions (half - life independent of initial concentration/pressure and volume).

Part b Answer: \(k = 6.93\times10^{-3}\ s^{-1}\)

Part c Answer: The initial rate in trial 1 is less than that in trial 2. The rate law for a first - order reaction is \(rate = k[cis - 2 - butene]\) (or \(rate\propto P_{cis - 2 - butene}\)). Trial 2 has a higher initial pressure (\(600\) torr vs \(300\) torr in trial 1), so the initial rate is higher in trial 2.

Part d Answer: The half - life in trial 4 is less than in trial 1 because trial 4 has a higher temperature (\(365\ K\) vs \(350\ K\) in trial 1). According to the Arrhenius equation, higher temperature increases the rate constant \(k\) (more molecules have energy ≥ activation energy). Since \(t_{1/2}=\frac{\ln2}{k}\) for a first - order reaction, a larger \(k\) leads to a smaller \(t_{1/2}\).