Question

At 25 oC, hydrogen iodide breaks down very slowly to hydrogen gas and iodine vapor with a rate constant of 2.4 x 10-21L/mol.s. If 0.0100 mol of HI(g) is placed into a 1.0 L container at 25 oC, how long will it take for the concentration of HI to reach 0.00900 mol/L?

Answer 1

Answer:

$\large \boxed{4.6 \times 10^{21}\text{ s}}$

Explanation:

Whenever a question asks you, "What is the concentration after a given time?" or something like that, you must use the appropriate integrated rate law expression.

The reaction is 2nd order, because the units of k are L·mol⁻¹s⁻¹.

The integrated rate law for a second-order reaction is  

$\dfrac{1}{\text{[A]}} =\dfrac{1}{\text{[A]}_{0}}+ kt$

Data:

   k = 2.4 × 10⁻²¹ L·mol⁻¹s⁻¹

[A]₀ = 0.0100     mol·L⁻¹

[A] = 0.009 00 mol·L⁻¹

Calculation :

$\begin{array}{rcl}\dfrac{1}{\text{[A]}} & = & \dfrac{1}{\text{[A]}_{0}}+ kt\\\\\dfrac{1}{0.00900 }& = & \dfrac{1}{0.0100} + 2.4 \times 10^{-21} \, t\\\\111.1&=& 100.0 + 2.4 \times 10^{-21} \, t\\\\11.1& = & 2.4 \times 10^{-21} \, t\\t & = & \dfrac{11.1}{ 2.4 \times 10^{-21}}\\\\& = & \mathbf{4.6 \times 10^{21}}\textbf{ s}\\\end{array}\\\text{It will take \large \boxed{\mathbf{4.6 \times 10^{21}}\textbf{ s}} for the HI to decompose}$