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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?

User Doll
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1 Answer

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Answer:


\large \boxed{4.6 * 10^(21)\text{ s}}

Step-by-step 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


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

User Mickyjtwin
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