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Chapter 11: Problem 51

The rate law for the decomposition of phosphine \(\left(\mathrm{PH}_{3}\right)\) is $$\text { Rate }=-\frac{\Delta\left[\mathrm{PH}_{3}\right]}{\Delta t}=k\left[\mathrm{PH}_{3}\right]$$ It takes \(120 .\) s for \(1.00 M\) PH \(_{3}\) to decrease to 0.250 M. How much time is required for \(2.00\space \mathrm{M} \mathrm{PH}_{3}\) to decrease to a concentration of \(0.350\space \mathrm{M} ?\)

Short Answer

Expert verified
It takes approximately 200 seconds for a 2.00 M concentration of PH3 to decrease to a concentration of 0.350 M.

Step by step solution

01

Determine the rate constant (k) using the given data

We can use the integrated rate law for a first-order reaction to solve this exercise. The integrated rate law for a first-order reaction is given by: \[ \ln\left(\frac{[\mathrm{PH}_{3}]_{0}}{[\mathrm{PH}_{3}]_{t}}\right) = kt \] where \([PH_3]_0\) is the initial concentration, \([PH_3]_t\) is the concentration at time t, k is the rate constant, and t is the time. We can use the data given in the exercise to calculate the rate constant (k). The initial concentration is 1.00 M, it decreases to 0.250 M after 120 s. Plug in the values and solve for k: \[ \ln\left(\frac{1.00\: \mathrm{M}}{0.250\: \mathrm{M}}\right) = k(120 \:s) \]
02

Calculating k

Calculate k. \[ k = \frac{\ln\left(\frac{1.00\: \mathrm{M}}{0.250\: \mathrm{M}}\right)}{120 \:s} \] \[ k \approx 0.00577\: \mathrm{s}^{-1} \]
03

Use the rate constant to find the time for the new initial concentration

Now that we have the rate constant k, we can use it to find the time required for a 2.00 M concentration of PH3 to decrease to 0.350 M. Plug in the values and solve for t: \[ \ln\left(\frac{2.00\: \mathrm{M}}{0.350\: \mathrm{M}}\right) = (0.00577 \: \mathrm{s}^{-1})t\]
04

Calculating t

Calculate t. \[ t = \frac{\ln\left(\frac{2.00\: \mathrm{M}}{0.350\: \mathrm{M}}\right)}{0.00577 \: \mathrm{s}^{-1}} \] \[ t \approx 199.99 \: \mathrm{s} \] The time required for a 2.00 M concentration of PH3 to decrease to a concentration of 0.350 M is approximately 200 seconds.

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Most popular questions from this chapter

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