For each of the questions, four choices have been provided. Select the correct alternative. For a particular reaction, \(A+B \rightarrow C\) was studied at \(25^{\circ} \mathrm{C}\). The following results are obtained: $$ \begin{array}{|c|c|c|} \hline \begin{array}{c} {[\mathrm{A}]} \\ \text { mole } / \mathrm{I} \end{array} & \begin{array}{c} {[\mathrm{B}]} \\ \text { moles } / \mathrm{I} \end{array} & \begin{array}{c} \text { Rate } \\ \left(\mathrm{mole} \mathrm{I}^{-1} \mathrm{~s}^{-1}\right) \end{array} \\ \hline 9 \times 10^{-5} & 1.5 \times 10^{-2} & 0.06 \\ \hline 9 \times 10^{-5} & 3 \times 10^{-3} & 0.012 \\ \hline 3 \times 10^{-5} & 3 \times 10^{-3} & 0.004 \\ \hline 6 \times 10^{-5} & \mathrm{x} & 0.024 \\ \hline \end{array} $$ Then the value of \(\mathrm{x}\) is ________ (a) \(6 \times 10^{-3} \mathrm{~m} \ell^{-1}\) (b) \(3 \times 10^{-3} \mathrm{~m} \ell^{-1}\) (c) \(4.5 \times 10^{-3} \mathrm{~m} \ell^{-1}\) (d) \(9 \times 10^{-3} \mathrm{~m} \ell^{-1}\)

Step by step solution

01

Determine the order of the reaction with respect to A

We start by comparing the first and second row of the data. The concentration of A is the same in both cases, while the concentration of B is 5 times smaller in the second row. Observe how the rate changes: $$\frac{0.012}{0.06} = \frac{1}{5}$$ This implies that the reaction is first-order with respect to B.

02

Determine the order of the reaction with respect to B

Now, we will compare the second and third row of the data. The concentration of B is the same in both cases, while the concentration of A is 3 times smaller in the third row. Observe how the rate changes: $$\frac{0.004}{0.012} = \frac{1}{3}$$ This implies that the reaction is first-order with respect to A.

03

Apply the rate law to the reaction

Since the reaction is first-order with respect to both A and B, the rate law is: $$\text{Rate} = k[\mathrm{A}][\mathrm{B}]$$ Where k is the rate constant.

04

Calculate the rate constant (k) using the first row of data

Use the rate, concentration of A, and concentration of B from the first row of data to calculate k: $$k = \frac{\text{Rate}}{[\mathrm{A}][\mathrm{B}]} = \frac{0.06}{(9 \times 10^{-5})(1.5 \times 10^{-2})}$$ $$k = 0.44444$$

05

Find the concentration x using the last row of data

Now, we will use the rate and concentration of A from the last row of the data, and the rate constant k, to solve for x (concentration of B): $$0.024 = 0.44444(6 \times 10^{-5})[\mathrm{x}]$$ $$[\mathrm{x}] = \frac{0.024}{(0.44444)(6 \times 10^{-5})}$$ $$[\mathrm{x}] = 9 \times 10^{-3}\ \mathrm{m\ell^{-1}}$$ The correct answer is (d) \(9 \times 10^{-3}\ \mathrm{m\ell^{-1}}\).

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