Question

The hypothetical reaction $$ \mathrm{X}(g)+\frac{1}{2} \mathrm{Y}(g) \longrightarrow \text { products } $$ is first-order in \(\mathrm{X}\) and second-order in \(\mathrm{Y}\). The rate of the reaction is \(0.00389 \mathrm{~mol} / \mathrm{L} \cdot \min\) when \([\mathrm{X}]\) is \(0.150 \mathrm{M}\) and \([\mathrm{Y}]\) is \(0.0800 \mathrm{M}\) (a) What is the value for \(k\) ? (b) At what concentration of [Y] is the rate \(0.00948 \mathrm{~mol} /\) \(\mathrm{L} \cdot \min\) and \([\mathrm{X}]\) is \(0.0441 \mathrm{M} ?\) (c) At what concentration of \([\mathrm{X}]\) is the rate \(0.0124 \mathrm{~mol} /\) \(\mathrm{L} \cdot \min\) and \([\mathrm{Y}]=2[\mathrm{X}] ?\)

Short answer

Question: Find the rate constant, \(k\), the concentration of \(\mathrm{Y}\) at a given rate and \(\mathrm{X}\) concentration, and the concentration of \(\mathrm{X}\) at a given rate and \(\mathrm{Y}\) concentration for a chemical reaction given by: \(\mathrm{X}(g)+\frac{1}{2} \mathrm{Y}(g) \longrightarrow \text { products }\), which is first-order in \(\mathrm{X}\) and second-order in \(\mathrm{Y}\). Answer: The rate constant, \(k\), is \(16.2 \mathrm{L^2\cdot mol^{-2}\cdot min^{-1}}\). The concentration of \(\mathrm{Y}\) at the given rate and \(\mathrm{X}\) concentration is \(0.110\, \mathrm{M}\). The concentration of \(\mathrm{X}\) at the given rate and \(\mathrm{Y}\) concentration is \(0.0576\, \mathrm{M}\).

Step-by-step solution

Write the rate law equation

Since the reaction is first-order in \(\mathrm{X}\) and second-order in \(\mathrm{Y}\), the rate law equation is given by: $$ \text{Rate} = k [\mathrm{X}]^1 [\mathrm{Y}]^2 $$ where \(k\) is the rate constant and \([\mathrm{X}]\) and \([\mathrm{Y}]\) represent the concentration of \(\mathrm{X}\) and \(\mathrm{Y}\).

Solve for k (part a)

We are given the rate, concentration of \(\mathrm{X}\), and concentration of \(\mathrm{Y}\), so we can substitute these values into the rate law equation and solve for \(k\): $$ 0.00389 \mathrm{~mol/L\cdot min} = k (0.150 \mathrm{M})^1 (0.0800 \mathrm{M})^2 $$ Solve for \(k\): $$ k = \frac{0.00389 \mathrm{~mol/L\cdot min}}{(0.150 \mathrm{M})(0.0800 \mathrm{M})^2} = 16.2 \mathrm{L^2\cdot mol^{-2}\cdot min^{-1}} $$

Solve for the concentration of Y (part b)

We are now given the rate and concentration of \(\mathrm{X}\), and we need to find the concentration of \(\mathrm{Y}\). We can substitute the known values and the calculated value of \(k\): $$ 0.00948 \mathrm{~mol/L\cdot min} = (16.2 \mathrm{L^2\cdot mol^{-2}\cdot min^{-1}})(0.0441 \mathrm{M})([\mathrm{Y}]^2) $$ Solve for \([\mathrm{Y}]\): $$ [\mathrm{Y}]^2 = \frac{0.00948 \mathrm{~mol/L\cdot min}}{(16.2 \mathrm{L^2\cdot mol^{-2}\cdot min^{-1}})(0.0441 \mathrm{M})} $$ $$ [\mathrm{Y}]^2 = 0.0121 \mathrm{M^2} $$ $$ [\mathrm{Y}] = \sqrt{0.0121 \mathrm{M^2}} = 0.110\, \mathrm{M} $$

Solve for the concentration of X (part c)

We are given the rate and the relationship between the concentrations of \(\mathrm{X}\) and \(\mathrm{Y} ([\mathrm{Y}] = 2 [\mathrm{X}])\). Substitute the rate and the calculated value of \(k\) in the rate law equation: $$ 0.0124 \mathrm{~mol/L\cdot min} = (16.2 \mathrm{L^2\cdot mol^{-2}\cdot min^{-1}})[\mathrm{X}][(2[\mathrm{X}])^2] $$ Solve for \([\mathrm{X}]\): $$ 0.0124 \mathrm{~mol/L\cdot min} = 16.2 \mathrm{L^2\cdot mol^{-2}\cdot min^{-1}}(\,4[\mathrm{X}]^3) $$ $$ [\mathrm{X}]^3 = \frac{0.0124 \mathrm{~mol/L\cdot min}}{(16.2 \mathrm{L^2\cdot mol^{-2}\cdot min^{-1}})(4)} = 0.1912 \times 10^{-3} \mathrm{M^3} $$ $$ [\mathrm{X}] = \sqrt[3]{0.1912 \times 10^{-3} \mathrm{M^3}} = 0.0576\, \mathrm{M} $$ In summary, we found that (a) the rate constant \(k = 16.2 \mathrm{L^2\cdot mol^{-2}\cdot min^{-1}}\), (b) the concentration of \(\mathrm{Y}\) at the given rate and \(\mathrm{X}\) concentration is \(0.110\, \mathrm{M}\), and (c) the concentration of \(\mathrm{X}\) at the given rate and \(\mathrm{Y}\) concentration is \(0.0576\, \mathrm{M}\).