Question

Reaction: $2A + B \rightarrow 3C$

Given:

1. The reaction is first-order with respect to $[A]$ and zero-order with respect to $[B]$.
2. Rate constant ($k$) = $0.02 \, \text{s}^{-1}$ (for the disappearance of $A$).
3. Initial concentrations: $[A]_0 = 0.5 \, \text{M}$, $[B]_0 = 0.2 \, \text{M}$.

Questions:

1. Calculate the time (in seconds) for $[A]$ to decrease to $0.1 \, \text{M}$.
2. Determine the instantaneous rate of formation of $C$ when $[A] = 0.2 \, \text{M}$.
3. Find the half-life of $A$ if $[A]_0 = 1.0 \, \text{M}$ and $[B]_0 = 0.1 \, \text{M}$.

Answer 1

1. Time for $[A]$ to decrease to $0.1 \, \text{M}$: $80.47 \, \text{s}$
2. Instantaneous rate of formation of $C$: $0.006 \, \text{M s}^{-1}$
3. Half-life of $A$: Not achievable

Answer 2

The problem involves applying the concepts of chemical kinetics, including rate laws, integrated rate laws, stoichiometry, and limiting reactants.

Given Information:

• Reaction: $2A + B \rightarrow 3C$
• Rate Law: First-order with respect to $[A]$ and zero-order with respect to $[B]$. This means the rate law is $\text{Rate} = k[A]$. For the disappearance of $A$, $-\frac{d[A]}{dt} = k[A]$.
• Rate constant (k): $0.02 \, \text{s}^{-1}$ (for the disappearance of $A$). The unit $s^{-1}$ confirms it's a first-order reaction.
• Initial concentrations: $[A]_0 = 0.5 \, \text{M}$, $[B]_0 = 0.2 \, \text{M}$.

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1. Calculate the time (in seconds) for $[A]$ to decrease to $0.1 \, \text{M}$.

Since the reaction is first-order with respect to $A$, the integrated rate law is: $\ln\left(\frac{[A]_t}{[A]_0}\right) = -kt$ We are given:

• $[A]_0 = 0.5 \, \text{M}$
• $[A]_t = 0.1 \, \text{M}$
• $k = 0.02 \, \text{s}^{-1}$

Substitute the values into the equation: $\ln\left(\frac{0.1 \, \text{M}}{0.5 \, \text{M}}\right) = -(0.02 \, \text{s}^{-1})t$ $\ln(0.2) = -0.02t$ $-1.6094 = -0.02t$ $t = \frac{1.6094}{0.02}$ $t = 80.47 \, \text{s}$

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2. Determine the instantaneous rate of formation of $C$ when $[A] = 0.2 \, \text{M}$.

First, calculate the instantaneous rate of disappearance of $A$ using the rate law: $-\frac{d[A]}{dt} = k[A]$ When $[A] = 0.2 \, \text{M}$ and $k = 0.02 \, \text{s}^{-1}$: $-\frac{d[A]}{dt} = (0.02 \, \text{s}^{-1})(0.2 \, \text{M})$ $-\frac{d[A]}{dt} = 0.004 \, \text{M s}^{-1}$

Next, relate the rate of disappearance of $A$ to the rate of formation of $C$ using the stoichiometry of the reaction $2A + B \rightarrow 3C$. The general rate expression based on stoichiometry is: $\text{Rate} = -\frac{1}{2}\frac{d[A]}{dt} = +\frac{1}{3}\frac{d[C]}{dt}$ We need to find $+\frac{d[C]}{dt}$: $\frac{d[C]}{dt} = -\frac{3}{2}\frac{d[A]}{dt}$ Substitute the calculated rate of disappearance of $A$ (which is a positive value): $\frac{d[C]}{dt} = \frac{3}{2} \times (0.004 \, \text{M s}^{-1})$ $\frac{d[C]}{dt} = 1.5 \times 0.004 \, \text{M s}^{-1}$ $\frac{d[C]}{dt} = 0.006 \, \text{M s}^{-1}$

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3. Find the half-life of $A$ if $[A]_0 = 1.0 \, \text{M}$ and $[B]_0 = 0.1 \, \text{M}$.

For a first-order reaction, the half-life ($t_{1/2}$) is given by: $t_{1/2} = \frac{\ln(2)}{k}$ Given $k = 0.02 \, \text{s}^{-1}$: $t_{1/2} = \frac{0.693}{0.02 \, \text{s}^{-1}}$ $t_{1/2} = 34.65 \, \text{s}$

However, we must check if the reaction can proceed long enough for $A$ to reach its half-life concentration given the initial amounts of reactants. For $[A]_0 = 1.0 \, \text{M}$, the half-life means $[A]$ decreases to $[A]_0/2 = 1.0 \, \text{M} / 2 = 0.5 \, \text{M}$. The amount of $A$ that needs to be consumed is $\Delta[A] = 1.0 \, \text{M} - 0.5 \, \text{M} = 0.5 \, \text{M}$.

From the stoichiometry $2A + B \rightarrow 3C$, 2 moles of $A$ react with 1 mole of $B$. To consume $0.5 \, \text{M}$ of $A$, the amount of $B$ required is: $\text{Required } \Delta[B] = \frac{1}{2} \times \Delta[A] = \frac{1}{2} \times 0.5 \, \text{M} = 0.25 \, \text{M}$ The initial concentration of $B$ available is $[B]_0 = 0.1 \, \text{M}$. Since the required amount of $B$ ($0.25 \, \text{M}$) is greater than the available amount of $B$ ($0.1 \, \text{M}$), $B$ is the limiting reactant. The reaction will stop when all of $B$ is consumed.

When $0.1 \, \text{M}$ of $B$ is consumed, the amount of $A$ consumed will be: $\Delta[A] = 2 \times \Delta[B] = 2 \times 0.1 \, \text{M} = 0.2 \, \text{M}$ The concentration of $A$ remaining at this point will be: $[A]_{final} = [A]_0 - \Delta[A] = 1.0 \, \text{M} - 0.2 \, \text{M} = 0.8 \, \text{M}$ Since the final concentration of $A$ ($0.8 \, \text{M}$) is greater than its half-life concentration ($0.5 \, \text{M}$), the half-life of $A$ is not achievable under these conditions because $B$ runs out first.

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Explanation of the solution:

1. Time for [A] to decrease: Use the integrated rate law for a first-order reaction: $t = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]_t}\right)$. Substitute the given initial and final concentrations of A and the rate constant to find the time.
2. Instantaneous rate of formation of C: First, calculate the instantaneous rate of disappearance of A using the given rate law $-\frac{d[A]}{dt} = k[A]$. Then, use the stoichiometric relationship from the balanced chemical equation $-\frac{1}{2}\frac{d[A]}{dt} = +\frac{1}{3}\frac{d[C]}{dt}$ to find the rate of formation of C.
3. Half-life of A: Calculate the theoretical half-life for a first-order reaction using $t_{1/2} = \frac{\ln(2)}{k}$. Then, check if the reaction can actually proceed to the half-life concentration of A by determining if the limiting reactant (B) runs out before A reaches half its initial concentration. If B is exhausted first, A's half-life is not achievable.