Question

18\. Complete the following table for the reaction below. It is first-order in both \(\mathrm{X}\) and \(\mathrm{Y}\). \(2 \mathrm{X}(g)+\mathrm{Y}(g) \longrightarrow\) products $$ \begin{array}{lcccc} \hline & {[\mathrm{X}]} & {[\mathrm{Y}]} & \mathrm{k}(\mathrm{L} / \mathrm{mol} \cdot \mathrm{h}) & \text { rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{h}) \\ \hline \text { (a) } & 0.100 & 0.400 & 1.89 & \\ \text { (b) } & 0.600 & & 0.884 & 0.159 \\ \text { (c) } & & 0.250 & 13.4 & 0.0479 \\ \text { (d) } & 0.600 & 0.233 & & 0.00112 \\ \hline \end{array} $$

Short answer

\\\ \text { (b) } & 0.600 & ? & 0.884 & 0.159 \\\ \text { (c) } & ? & 0.250 & 13.4 & 0.0479 \\\ \text { (d) } & 0.600 & 0.233 & ? & 0.00112 \\\ \hline \end{array} $$ Answer: The complete table with the missing values calculated is: $$ \begin{array}{lcccc} \hline & {[\mathrm{X}]} & {[\mathrm{Y}]} & \mathrm{k}(\mathrm{L} / \mathrm{mol} \cdot \mathrm{h}) & \text { rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{h}) \\\ \hline \text { (a) } & 0.100 & 0.400 & 1.89 & 0.0756 \\\ \text { (b) } & 0.600 & 0.300 & 0.884 & 0.159 \\\ \text { (c) } & 0.0141 & 0.250 & 13.4 & 0.0479 \\\ \text { (d) } & 0.600 & 0.233 & 0.00795 & 0.00112 \\\ \hline \end{array} $$

Step-by-step solution

Write the rate equation for the first-order reaction

The rate equation for this first-order reaction involving X and Y is given by: Rate = k [X] [Y] Now we will use the rate equation to find the missing values in each row.

Calculate missing value of a

In row (a), we are given [X], [Y], and k. We can use the rate equation to find the rate: Rate = (1.89 \(\frac{L}{mol\cdot h}\)) (0.100 mol/L) (0.400 mol/L) = 0.0756 \(\frac{mol}{L\cdot h}\)

Calculate missing value of b

In row (b), we are given [X], k, and the rate. We can use the rate equation to find [Y]: [Y] = \(\frac{Rate}{k [X]}\) = \(\frac{0.159 \frac{mol}{L \cdot h}}{0.884 \frac{L}{mol\cdot h} (0.600 \, mol/L)}\) = 0.3 mol/L

Calculate missing value of c

In row (c), we are given [Y], k, and the rate. We can use the rate equation to find [X]: [X] = \(\frac{Rate}{k [Y]}\) = \(\frac{0.0479 \frac{mol}{L \cdot h}}{13.4 \frac{L}{mol\cdot h} (0.250 \, mol/L)}\) = 0.0141 mol/L

Calculate missing value of d

In row (d), we are given [X], [Y], and the rate. We can use the rate equation to find k: k = \(\frac{Rate}{[X][Y]}\) = \(\frac{0.00112 \frac{mol}{L \cdot h}}{(0.600 \, mol/L) (0.233 \, mol/L)}\) = 0.00795 \(\frac{L}{mol\cdot h}\) So, the complete table looks like this: $$ \begin{array}{lcccc} \hline & {[\mathrm{X}]} & {[\mathrm{Y}]} & \mathrm{k}(\mathrm{L} / \mathrm{mol} \cdot \mathrm{h}) & \text { rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{h}) \\\ \hline \text { (a) } & 0.100 & 0.400 & 1.89 & 0.0756 \\\ \text { (b) } & 0.600 & 0.300 & 0.884 & 0.159 \\\ \text { (c) } & 0.0141 & 0.250 & 13.4 & 0.0479 \\\ \text { (d) } & 0.600 & 0.233 & 0.00795 & 0.00112 \\\ \hline \end{array} $$