Question

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

Given:

1. The reaction follows a complex rate law: Rate = k[A]²[B] when [B] is present
2. Pseudo-first-order conditions apply at low [B]
3. k = 0.04 M⁻²s⁻¹ at 350K
4. Initial concentrations: [A]₀ = 0.8 M, [B]₀ = 0.2 M

Questions:

1. Calculate the initial rate of disappearance of A
2. Determine the time required for [A] to decrease to 0.2 M under conditions where [B] remains approximately constant
3. Find the instantaneous rate of formation of C when [A] = 0.4 M (assuming [B] ≈ 0.15 M at that moment)
4. If the reaction were changed to zero-order with respect to A under certain conditions, what would be the half-life when [A]₀ = 0.6 M and k = 0.03 M/s

Answer 1

1. Initial rate of disappearance of A: $0.01536 \, M s^{-1}$
2. Time required for [A] to decrease to 0.2 M (constant [B]): $156.25 \, s$
3. Instantaneous rate of formation of C when [A] = 0.4 M ([B] = 0.15 M): $0.00096 \, M s^{-1}$
4. Half-life for zero-order A: $10 \, s$

Answer 2

The problem presents a chemical reaction and a complex rate law, along with specific conditions for different parts of the question. A significant inconsistency arises from B being a product in the stoichiometric equation ($3A \rightarrow 2B + C$) but appearing as a reactant in the rate law (Rate = $k[A]^2[B]$). Furthermore, its concentration behavior is specified in contradictory ways across different parts. We will proceed by strictly adhering to the conditions and values provided for each sub-question, acknowledging these inconsistencies.

Given Information:

• Reaction: $3A \rightarrow 2B + C$
• Rate Law: Rate = $k[A]^2[B]$ (This is the rate of reaction, $Rate = -\frac{1}{3}\frac{d[A]}{dt} = +\frac{1}{2}\frac{d[B]}{dt} = +\frac{d[C]}{dt}$)
• Rate Constant: $k = 0.04 \, M^{-2}s^{-1}$ at 350K
• Initial Concentrations: $[A]_0 = 0.8 \, M$, $[B]_0 = 0.2 \, M$

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1. Calculate the initial rate of disappearance of A

The initial rate of reaction is given by the rate law using initial concentrations: $\text{Initial Rate} = k[A]_0^2[B]_0$ Substitute the given values: $\text{Initial Rate} = (0.04 \, M^{-2}s^{-1})(0.8 \, M)^2(0.2 \, M)$ $\text{Initial Rate} = (0.04)(0.64)(0.2) \, M s^{-1}$ $\text{Initial Rate} = 0.00512 \, M s^{-1}$ The rate of disappearance of A is related to the rate of reaction by the stoichiometry: $-\frac{d[A]}{dt} = 3 \times \text{Rate}$ $\text{Initial Rate of disappearance of A} = 3 \times 0.00512 \, M s^{-1}$ $\text{Initial Rate of disappearance of A} = 0.01536 \, M s^{-1}$

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2. Determine the time required for [A] to decrease to 0.2 M under conditions where [B] remains approximately constant

If [B] remains approximately constant, the rate law simplifies to a pseudo-second-order reaction with respect to A. We use the initial concentration of B for this calculation. The rate of disappearance of A is: $-\frac{d[A]}{dt} = 3 \times \text{Rate} = 3k[A]^2[B]$ Let $k' = 3k[B]$. Since [B] is constant (at $[B]_0 = 0.2 \, M$), $k'$ is a constant: $k' = 3 \times (0.04 \, M^{-2}s^{-1}) \times (0.2 \, M)$ $k' = 0.024 \, M^{-1}s^{-1}$ The integrated rate law for a second-order reaction (with respect to A) is: $\frac{1}{[A]_t} - \frac{1}{[A]_0} = k't$ We want to find $t$ when $[A]_t = 0.2 \, M$, with $[A]_0 = 0.8 \, M$: $\frac{1}{0.2 \, M} - \frac{1}{0.8 \, M} = (0.024 \, M^{-1}s^{-1})t$ $5 \, M^{-1} - 1.25 \, M^{-1} = (0.024 \, M^{-1}s^{-1})t$ $3.75 \, M^{-1} = (0.024 \, M^{-1}s^{-1})t$ $t = \frac{3.75}{0.024} \, s$ $t = 156.25 \, s$

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3. Find the instantaneous rate of formation of C when [A] = 0.4 M (assuming [B] ≈ 0.15 M at that moment)

The instantaneous rate of formation of C is equal to the rate of reaction ($+\frac{d[C]}{dt} = \text{Rate}$). We use the given instantaneous concentrations for [A] and [B]. $\text{Rate} = k[A]^2[B]$ Substitute the given values: $\text{Rate} = (0.04 \, M^{-2}s^{-1})(0.4 \, M)^2(0.15 \, M)$ $\text{Rate} = (0.04)(0.16)(0.15) \, M s^{-1}$ $\text{Rate} = 0.04 \times 0.024 \, M s^{-1}$ $\text{Rate} = 0.00096 \, M s^{-1}$ Therefore, the instantaneous rate of formation of C is $0.00096 \, M s^{-1}$.

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4. If the reaction were changed to zero-order with respect to A under certain conditions, what would be the half-life when [A]₀ = 0.6 M and k = 0.03 M/s?

For a zero-order reaction with respect to A, the rate of disappearance of A is constant: $-\frac{d[A]}{dt} = k_{zero}$ The given rate constant $k = 0.03 \, M/s$ has units consistent with a zero-order rate of disappearance of A. The integrated rate law for a zero-order reaction is: $[A]_t = [A]_0 - k_{zero}t$ The half-life ($t_{1/2}$) is the time when $[A]_t = [A]_0/2$: $\frac{[A]_0}{2} = [A]_0 - k_{zero}t_{1/2}$ $k_{zero}t_{1/2} = [A]_0 - \frac{[A]_0}{2}$ $k_{zero}t_{1/2} = \frac{[A]_0}{2}$ $t_{1/2} = \frac{[A]_0}{2k_{zero}}$ Substitute the given values: $t_{1/2} = \frac{0.6 \, M}{2 \times (0.03 \, M/s)}$ $t_{1/2} = \frac{0.6}{0.06} \, s$ $t_{1/2} = 10 \, s$

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Explanation of the solution:

1. Initial Rate of Disappearance of A: Calculated the initial reaction rate using the given rate law and initial concentrations. Then, multiplied by the stoichiometric coefficient of A (3) to find its disappearance rate.
2. Time for [A] to Decrease (Constant [B]): Assumed [B] remains constant at its initial value as specified. The rate law becomes pseudo-second-order with respect to A. Applied the integrated rate law for a second-order reaction to find the time.
3. Instantaneous Rate of Formation of C: Used the given instantaneous concentrations of [A] and [B] directly in the rate law. The rate of reaction equals the rate of formation of C.
4. Half-life for Zero-Order A: Applied the half-life formula for a zero-order reaction, $t_{1/2} = \frac{[A]_0}{2k}$, using the provided initial concentration and zero-order rate constant.