Rate of Reaction from Initial Rates Kinetic Data

To determine the rate of the reaction, we start by writing the general rate law for the reaction $\text{A} + \text{B} + \text{C} \rightarrow \text{Product}$:

$\text{Rate} = k [\text{A}]^x [\text{B}]^y [\text{C}]^z$

where $k$ is the rate constant, and $x$, $y$, and $z$ are the partial orders of the reaction with respect to reactants $\text{A}$, $\text{B}$, and $\text{C}$, respectively.

Step 1: Determination of the Order with respect to Reactant B ($y$)

Comparing Experiment 1 and Experiment 2:

• Concentrations of $\text{A}$ and $\text{C}$ are kept constant: $[\text{A}] = 0.2\text{ mol dm}^{-3}$ and $[\text{C}] = 0.1\text{ mol dm}^{-3}$.
• Concentration of $\text{B}$ is doubled from $0.1\text{ mol dm}^{-3}$ to $0.2\text{ mol dm}^{-3}$.
• The rate of reaction remains unchanged at $6.0 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}$.

$\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{6.0 \times 10^{-5}}{6.0 \times 10^{-5}} = \left(\frac{0.2}{0.1}\right)^y$ $1 = 2^y \implies y = 0$

Thus, the reaction is zeroth-order with respect to $\text{B}$.

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Step 2: Determination of the Order with respect to Reactant C ($z$)

Comparing Experiment 1 and Experiment 3:

• Concentrations of $\text{A}$ and $\text{B}$ are kept constant: $[\text{A}] = 0.2\text{ mol dm}^{-3}$ and $[\text{B}] = 0.1\text{ mol dm}^{-3}$.
• Concentration of $\text{C}$ is doubled from $0.1\text{ mol dm}^{-3}$ to $0.2\text{ mol dm}^{-3}$.
• The rate of reaction doubles from $6.0 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}$ to $1.2 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}$.

$\frac{\text{Rate}_3}{\text{Rate}_1} = \frac{1.2 \times 10^{-4}}{6.0 \times 10^{-5}} = \left(\frac{0.2}{0.1}\right)^z$ $2 = 2^z \implies z = 1$

Thus, the reaction is first-order with respect to $\text{C}$.

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Step 3: Determination of the Order with respect to Reactant A ($x$)

Comparing Experiment 1 and Experiment 4:

• Concentrations of $\text{B}$ and $\text{C}$ are kept constant: $[\text{B}] = 0.1\text{ mol dm}^{-3}$ and $[\text{C}] = 0.1\text{ mol dm}^{-3}$.
• Concentration of $\text{A}$ is increased by a factor of $\frac{0.3}{0.2} = 1.5$.
• The rate of reaction increases from $6.0 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}$ to $9.0 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}$ (a factor of $1.5$).

$\frac{\text{Rate}_4}{\text{Rate}_1} = \frac{9.0 \times 10^{-5}}{6.0 \times 10^{-5}} = \left(\frac{0.3}{0.2}\right)^x$ $1.5 = (1.5)^x \implies x = 1$

Thus, the reaction is first-order with respect to $\text{A}$.

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Step 4: Calculation of the Rate Constant ($k$)

The overall rate law is: $\text{Rate} = k [\text{A}] [\text{C}]$

Using data from Experiment 1: $6.0 \times 10^{-5} = k \times (0.2) \times (0.1)$ $k = \frac{6.0 \times 10^{-5}}{0.02} = 3.0 \times 10^{-3}\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}$

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Step 5: Calculation of the New Rate and Value of $Y$

For the given concentrations:

• $[\text{A}] = 0.15\text{ mol dm}^{-3}$
• $[\text{B}] = 0.25\text{ mol dm}^{-3}$
• $[\text{C}] = 0.15\text{ mol dm}^{-3}$

The new rate of reaction is: $\text{Rate} = k [\text{A}] [\text{C}]$ $\text{Rate} = (3.0 \times 10^{-3}) \times (0.15) \times (0.15)$ $\text{Rate} = 3.0 \times 10^{-3} \times 0.0225 = 6.75 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}$

Given that the reaction rate is expressed as $Y \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}$, we get: $Y = 6.75$