The reaction $$\mathrm{ICl}(g)+\frac{1}{2} \mathrm{H}_{2}(g) \longrightarrow \frac{1}{2} \mathrm{I}_{2}(g)+\mathrm{HCl}(g)$$ is first-order in both reactants. The rate of the reaction is \(4.89 \times 10^{-5} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) when the ICl concentration is \(0.100 M\) and that of the hydrogen gas is \(0.030 \mathrm{M}\) (a) What is the value of \(k\) ? (b) At what concentration of hydrogen is the rate \(5.00 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) and \([\mathrm{ICl}]=0.233 \mathrm{M?}\) (c) At what concentration of iodine chloride is the rate \(0.0934 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) if the hydrogen concentration is three times that of ICl?

Step by step solution

01

Write the rate law equation

Since the reaction is first-order with respect to both reactants, the rate law equation is given by: $$rate = k[\mathrm{ICl}][\mathrm{H}_{2}]$$

02

Use the given information to find the value of \(k\)

We are given the rate as \(4.89 \times 10^{-5} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) when the concentration of ICl is \([\mathrm{ICl}] = 0.100 M\) and the concentration of hydrogen gas is \([\mathrm{H}_{2}] = 0.030 M\). Substituting these values into the rate law equation, we get: $$4.89 \times 10^{-5} = k(0.100)(0.030)$$ Now solve for \(k\): $$k = \frac{4.89 \times 10^{-5}}{(0.100)(0.030)} = 1.63 \times 10^{-3}\ \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}$$ (b) Find the concentration of hydrogen gas when the rate is \(5.00 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) and the concentration of ICl is \(0.233 M\)

03

Replace the known values into the rate law equation

We are given the rate, the value of \(k\) from part (a), and the concentration of ICl. We need to find the concentration of hydrogen gas \([\mathrm{H}_{2}]\). Use the rate law equation: $$5.00 \times 10^{-4} = (1.63 \times 10^{-3})(0.233)[\mathrm{H}_{2}]$$

04

Solve for the concentration of hydrogen gas

Rearrange the equation to solve for \([\mathrm{H}_{2}]\): $$[\mathrm{H}_{2}] = \frac{5.00 \times 10^{-4}}{(1.63 \times 10^{-3})(0.233)} = 1.321 \mathrm{M}$$ So the concentration of hydrogen gas at this rate is \(1.321 M\). (c) Find the concentration of ICl when the rate is \(0.0934 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) and the hydrogen concentration is three times that of ICl

05

Write the rate law equation with given conditions

We are given the rate and the relationship between the concentrations of ICl and hydrogen gas. Let \(x\) be the concentration of ICl. Then, the concentration of hydrogen gas is \(3x\). Use the rate law equation: $$0.0934 = (1.63 \times 10^{-3})(x)(3x)$$

06

Solve for the concentration of ICl

Rearrange the equation and solve for \(x\): $$x^2 = \frac{0.0934}{(1.63 \times 10^{-3})(3)} \Rightarrow x = \sqrt{\frac{0.0934}{(1.63 \times 10^{-3})(3)}} = 2.341 \mathrm{M}$$ So the concentration of iodine chloride is \(2.341 M\) under these conditions.

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