Question 5b733

Dec 24, 2014

$t = 0.515 s$ (for A) and $t = 1.74 s$ (for B).

The general form of a first order reaction is

$a A \to P R O D U C T S$, with the reaction rate being

$r a t e = k \cdot \left[A\right]$.

The equation we'll use for this problem is

$\ln \left(\frac{\left[A\right]}{A} _ \circ\right) = - k t$, where

${\left[A\right]}_{0}$ - the initial concentration of $A$;
$\left[A\right]$ - the final concentration of $A$;
$k$ - the rate constant - in your case equal to $0.693 {s}^{-} 1$
$t$ - the reaction time;

Now, 30% decomposition means that the final concentration of $A$ will equal 70% of the initial concentration - 30% consumed, 70% remaining. This can be written as

$\left[A\right] = \frac{100 - 30}{100} \cdot {\left[A\right]}_{0} = \frac{7}{10} \cdot {\left[A\right]}_{0}$

The equation becomes

$\ln \left(\frac{\frac{7}{10} \cdot {\left[A\right]}_{0}}{A} _ 0\right) = - k \cdot t$

$\ln \left(\frac{7}{10}\right) = - 0.693 \cdot t \to t = \ln \frac{0.7}{- 0.693 {s}^{-} 1} = 0.515 s$

The same approach can be used for when 70%# decomposition occurs:

$\left[A\right] = \frac{100 - 70}{100} \cdot {\left[A\right]}_{0} = \frac{3}{10} \cdot {\left[A\right]}_{0}$

$\ln \left(\frac{\frac{3}{10} \cdot {\left[A\right]}_{0}}{A} _ 0\right) = - k t \to t = \ln \frac{0.3}{- 0.693 {s}^{-} 1} = 1.74 s$