Question

For the reaction \(\mathrm{A}(g)+\mathrm{B}(g)+\mathrm{C}(g) \longrightarrow \mathrm{D}(g)\) the following data were obtained at constant temperature: $$ \begin{array}{ccccc} \text { Expt } & \begin{array}{c} \text { Initial Rate } \\ (\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) \end{array} & \begin{array}{c} \text { Initial [A] } \\ (\mathrm{mol} / \mathrm{L}) \end{array} & \begin{array}{c} \text { Initial [B] } \\ (\mathrm{mol} / \mathrm{L}) \end{array} & \begin{array}{c} \text { Initial [C] } \\ (\mathrm{mol} / \mathrm{L}) \end{array} \\ \hline 1 & 6.25 \times 10^{-3} & 0.0500 & 0.0500 & 0.0100 \\ 2 & 1.25 \times 10^{-2} & 0.1000 & 0.0500 & 0.0100 \\ 3 & 5.00 \times 10^{-2} & 0.1000 & 0.1000 & 0.0100 \\ 4 & 6.25 \times 10^{-3} & 0.0500 & 0.0500 & 0.0200 \end{array} $$ (a) What is the order with respect to each reactant? (b) Write the rate law. (c) Calculate \(k\) (using the data from Expt 1 ). (d) Using the value of \(k\) calculated in part (c), calculate the rate when \([\mathrm{A}]=\) \(0.2000 \mathrm{~mol} / \mathrm{L},[\mathrm{B}]=0.1500 \mathrm{~mol} / \mathrm{L},\) and \([\mathrm{C}]=0.0300 \mathrm{~mol} / \mathrm{L}\).

Short answer

The reaction is first order with respect to [A], second order with respect to [B], and zero order with respect to [C]. The rate law is \[ \text{Rate} = k[A][B]^2 \], where k = 50. The rate is 0.225 mol L^{-1} s^{-1} for the given concentrations.

Step-by-step solution

Determine Order with Respect to [A]

Compare experiments 1 and 2 where [B] and [C] are constant. The initial rate increases from \[6.25 \times 10^{-3}\] to \[1.25 \times 10^{-2}\] when [A] doubles from 0.050 M to 0.100 M. The rate also doubles. Thus, the reaction is first order with respect to A.

Determine Order with Respect to [B]

Compare experiments 2 and 3 where [A] and [C] are constant. The initial rate increases from \[1.25 \times 10^{-2}\] to \[5.00 \times 10^{-2}\] when [B] doubles from 0.050 M to 0.100 M. The rate quadruples. Thus, the reaction is second order with respect to B.

Determine Order with Respect to [C]

Compare experiments 1 and 4 where [A] and [B] are constant. The initial rate remains the same when [C] doubles from 0.010 M to 0.020 M. Thus, the reaction is zero order with respect to C.

Write the Rate Law

Given that the reaction is first order with respect to A, second order with respect to B, and zero order with respect to C, the rate law can be written as: \[ \text{Rate} = k[A][B]^2[C]^0 = k[A][B]^2 \]

Calculate the Rate Constant, k

Using the data from experiment 1: \[ \text{Rate} = k[A][B]^2 \] \[ 6.25 \times 10^{-3} = k(0.0500)(0.0500)^2 \] \[ 6.25 \times 10^{-3} = k(0.0500)(0.0025) \] \[ 6.25 \times 10^{-3} = k(0.000125) \] \[ k = \frac{6.25 \times 10^{-3}}{0.000125} = 50 \space \mathrm{L}^2 \space \mathrm{mol}^{-2} \space \mathrm{s}^{-1} \]

Calculate the Rate with Given Concentrations

Using the rate constant from step 5 and given concentrations: \[ \text{Rate} = k[A][B]^2 \] \[ \text{Rate} = 50 (0.2000)(0.1500)^2 \] \[ \text{Rate} = 50 (0.2000)(0.0225) \] \[ \text{Rate} = 50 (0.0045) \] \[ \text{Rate} = 0.225 \space \mathrm{mol} \space \mathrm{L}^{-1} \space \mathrm{s}^{-1} \]

Key concepts

reaction order

To grasp the concept of 'reaction order', it's essential to understand that it defines how the rate of reaction depends on the concentration of each reactant. For the reaction \(\text{A}(g) + \text{B}(g) + \text{C}(g) \to \text{D}(g)\), we determine the order by observing how changes in reactant concentrations affect the reaction rate.
Let's break it down:

• **First Order Reaction:** If doubling the concentration of a reactant doubles the rate of reaction, the reaction is first order with respect to that reactant.
• **Second Order Reaction:** If doubling the concentration makes the rate quadruple, it's second order.
• **Zero Order Reaction:** If changing the concentration doesn't affect the rate, it's zero order.

Using these principles, by comparing experiments where the other reactants are kept constant, you can determine the reaction order for each reactant. For instance, for \[ \text{A} \], the rate doubled when its concentration doubled – indicating a first-order reaction. This analysis helps in creating the rate law which brings us to our next topic.

rate law

The 'rate law' is an equation that links the rate of reaction to the concentrations of reactants, each raised to a power corresponding to their order in the reaction. For the given reaction, the rate law derived from the orders found is:
\[ \text{Rate} = k[A][B]^2[C]^0 \]
Where:

• \(\text{[A]}\) and \(\text{[B]}\) are concentrations of reactants A and B, respectively.
• The exponents (1 for \(\text{[A]}\) and 2 for \(\text{[B]}\)) are the orders of the reaction with respect to these reactants.
• \(\text{k}\) is the rate constant.

The \[ \text{[C]}^0 \] term simplifies to 1 (since any number to the power of 0 is 1), showing that the rate is unaffected by changes in \(\text{[C]}\). This rate law provides a concise way to predict how altering concentrations impacts the reaction rate, which is crucial for controlling chemical processes.

rate constant

The 'rate constant' (k) is a proportionality factor in the rate law that is specific to a given reaction and temperature. It links the reaction rate to the concentrations of reactants. To find \(k\), use the rate law and experimental data. Using experiment 1 data:
\(\text{Rate} = k[A][B]^2\)
Inserting the given values:
\(\text{6.25 \times 10^{-3}} = k(0.0500)(0.0500)^2\)
Solve for \(k\):
\(k = \frac{6.25 \times 10^{-3}}{(0.0500 \times 0.0025)}\)
Resulting in \(k = 50 \mathrm{L}^2 \mathrm{mol}^{-2} \mathrm{s}^{-1} \).
Understanding \(k\)'s value and unit helps predict reaction speeds under different conditions. This information can be used to calculate the reaction rate at other conditions with the established rate law. For instance, given new concentrations, plug them into the rate law with the calculated \(k\):
\[ \text{Rate} = 50 \times (0.2000) \times (0.1500)^2 = 0.225 \textrm{ mol L}^{-1}s^{-1} \]. This integration of \(k\) with concentration values accurately forecasts reaction behaviors.