For the decomposition of (CH3)2O (species A) at 777 K, the time required for [A]0 to fall...

(a)

The reaction involved is as follows:

\(a \mathrm{~A} \rightarrow\) products

The rate law for the above reaction is:

$$ r=k[\mathrm{~A}]^{n} $$

The rate law for \(n \neq 1\) is given as follows:

$$ \begin{aligned} r &=k[\mathrm{~A}]^{n} \\ &=-\frac{1}{a} \frac{d[\mathrm{~A}]}{d t}=k[\mathrm{~A}]^{n} \end{aligned} $$

Or,

$$ \begin{aligned} \frac{d[\mathrm{~A}]}{[\mathrm{A}]^{n}} &=-a k d t \\ &=-k_{\mathrm{A}} d t \end{aligned} $$

Here, \(a k=k_{\mathrm{A}}\).

Separate the variables and integrate the above expression as follows:

$$ \int_{\left[A_{L}\right.}^{|A|}[A]^{-n} d[A]=-k_{A} \int_{0}^{t} d t \ldots \ldots .(1) $$

Set the integration limits of concentration from \([\mathrm{A}]_{0}\) to \([\mathrm{A}]\) and of time from 0 to \(t\). Here, \([\mathrm{A}]_{0}\) is \([\mathrm{A}]\) at \(t=0\)

Solve equation (1) as follows:

$$ \frac{[\mathrm{A}]^{-n+1}-[\mathrm{A}]_{0}^{n+1}}{-n+1}=-k_{\mathrm{A}} t $$

Multiply by \((1-n)[\mathrm{A}]_{0}^{n-1}\) on both the sides gives:

$$ \left(\frac{[\mathrm{A}]}{[\mathrm{A}]_{0}}\right)^{1-n}=1+[\mathrm{A}]_{0}^{n-1}(n-1) k_{\mathrm{A}} t \ldots \ldots(2) $$

Substitute \([\mathrm{A}]\) as \(0.69[\mathrm{~A}]_{0}\) and \(t\) as \(t_{a}\) (fractional life) as follows:

$$ [\mathrm{A}]_{0}^{w-1}(n-1) k_{\Lambda} t_{a}=\left(\frac{0.69[\mathrm{~A}]_{0}}{[\mathrm{~A}]_{0}}\right)^{1-n}-1 $$

Solve for \(t_{a}\) as follows:

$$ t_{a}=\frac{(0.69)^{1-n}-1}{[\mathrm{~A}]_{0}^{n-1}(n-1) k_{\mathrm{A}}} \ldots \ldots(3) $$

Take \(\log\) on both sides of equation (3) as follows:

$$ \log _{\models 0} t_{\alpha}=\log _{10} \frac{(0.69)^{1-n}-1}{(n-1) k_{\mathrm{A}}}-(n-1) \log _{10}[\mathrm{~A}]_{0} $$

The data is given as follows:

Plot the graph of and gives a straight line where slope is .

Compare the equation obtained from the graph with that of the straight line \(y=m x+c\).

It gives the slope \((m)\) as:

$$ m=1-n \ldots \ldots(4) $$

From the graph the value of \(m\) is \(-0.4426\).

Substitute \(m\) as \(-0.4426\) in equation (4) as follows:

$$ -0.4426=1-n $$

Solve for \(n\) as follows:

$$ \begin{aligned} n &=1.4426 \\ & \approx 1.5 \end{aligned} $$

(b)

The decomposition of \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O}\) requires time for \([\mathrm{A}]_{0}\) to fall to \(0.69[\mathrm{~A}]_{0}\).

The dimensionless parameter is given as:

$$ \alpha \equiv \frac{[\mathrm{A}]}{[\mathrm{A}]_{0}} $$

Here, \(\alpha\) denotes fraction of \(\mathrm{A}\) that is unreacted and \([\mathrm{A}]_{0}\) is \([\mathrm{A}]\) at \(t=0 . t_{a}\) is the fractional life.

Solve the value of \(\alpha\) as follows:

$$ \begin{aligned} \alpha &=\frac{0.69[\mathrm{~A}]_{0}}{[\mathrm{~A}]_{0}} \\ &=0.69 \end{aligned} $$

Substitute \(\alpha\) as \(0.69\) and \(n\) as \(1.5\) in equation (2) as follows:

$$ \begin{aligned} &(0.69)^{1-1.5}=1+[\mathrm{A}]_{0}^{1, S-1}(1.5-1) k_{\wedge} t_{\alpha} \\ &(0.69)^{-0.5}=1+[\mathrm{A}]_{0}^{\frac{1}{2}}(0.5) k_{\mathrm{A}} t_{\alpha} \\ &\frac{1.2038-1}{0.5}=[\mathrm{A}]_{0}^{\frac{1}{2}} k_{\wedge} t_{a} \end{aligned} $$

Solve for \(k_{\mathrm{A}}\) :

$$ \frac{0.408}{[\mathrm{~A}]_{0}^{\frac{1}{2}} t_{\alpha}}=k_{\mathrm{A}} \ldots \ldots(5) $$

Substitute for four pairs of the data given to calculate \(k_{\text {A }}\) for each pair as follows:

For pair 1-

Substitute \(t_{a}\) as \(590 \mathrm{~s}\) and \([\mathrm{A}]_{0}\) as \(0.00813 \mathrm{~mol} / \mathrm{dm}^{3}\) in equation (5) as follows:

$$ \frac{0.408}{\left(0.00813 \mathrm{~mol} / \mathrm{dm}^{3}\right)^{\frac{1}{2}} \times 590 \mathrm{~s}}=k_{\mathrm{A}_{1}} $$

Solve for \(k_{\mathrm{A}_{i}}\) :

$$ k_{A_{1}}=7.67 \times 10^{-3} \mathrm{dm}^{\frac{3}{2}} / \mathrm{mol}^{\frac{1}{2}}-\mathrm{s} $$

For pair 2-

Substitute \(t_{\alpha}\) as \(665 \mathrm{~s}\) and \([\mathrm{A}]_{0}\) as \(0.00644 \mathrm{~mol} / \mathrm{dm}^{3}\) in equation (5) as follows:

$$ \frac{0.408}{\left(0.00644 \mathrm{~mol} / \mathrm{dm}^{3}\right)^{\frac{1}{2}} \times 665 \mathrm{~s}}=k_{\mathrm{A}_{2}} $$

Solve for \(k_{A_{2}}\) :

$$ k_{A_{2}}=7.64 \times 10^{-3} \mathrm{dm}^{\frac{3}{2}} / \mathrm{mol}^{\frac{1}{2}}-\mathrm{s} $$

For pair 3-

Substitute \(t_{a}\) as \(900 \mathrm{~s}\) and \([\mathrm{A}]_{0}\) as \(0.00310 \mathrm{~mol} / \mathrm{dm}^{3}\) in equation (5) as follows:

$$ \frac{0.408}{\left(0.00310 \mathrm{~mol} / \mathrm{dm}^{3}\right)^{\frac{1}{2}} \times 900 \mathrm{~s}}=k_{A_{3}} $$

Solve for \(k_{\mathrm{A}}\) :

$$ k_{\mathrm{A}_{3}}=8.14 \times 10^{-3} \mathrm{dm}^{\frac{3}{2}} / \mathrm{mol}^{\frac{1}{2}}-\mathrm{s} $$

For pair 4-

Substitute \(t_{a}\) as \(1140 \mathrm{~s}\) and \([\mathrm{A}]_{0}\) as \(0.00188 \mathrm{~mol} / \mathrm{dm}^{3}\) in equation \((5)\) as follows:

$$ \frac{0.408}{\left(0.00188 \mathrm{~mol} / \mathrm{dm}^{3}\right)^{\frac{1}{2}} \times 1140 \mathrm{~s}}=k_{\Lambda_{4}} $$

Solve for \(k_{\mathrm{A}_{4}}:\)

$$ k_{\mathrm{A}_{4}}=8.25 \times 10^{-3} \mathrm{dm}^{\frac{3}{2}} / \mathrm{mol}^{\frac{1}{2}}-\mathrm{s} $$

Take average of rate constant values of all four pairs as \(k_{\mathrm{A}}\) :

$$ k_{\mathrm{A}}=\frac{k_{\mathrm{A}_{1}}+k_{\mathrm{A}_{2}}+k_{\mathrm{A}_{\mathrm{B}}}+k_{\mathrm{A}_{4}}}{4} $$

Substitute \(k_{A_{1}}\) as \(7.67 \times 10^{-3} \mathrm{dm}^{\frac{3}{2}} / \mathrm{mol}^{\frac{1}{2}}-\mathrm{s}^{,}, k_{\mathrm{A}_{2}}\) as \(7.64 \times 10^{-3} \mathrm{dm}^{\frac{3}{2}} / \mathrm{mol}^{\frac{1}{2}}-\mathrm{s}^{\prime}, k_{\mathrm{A}}\), as \(8.14 \times 10^{-3} \mathrm{dm}^{\frac{3}{2}} / \mathrm{mol}^{\frac{1}{2}}-\mathrm{s}\) and \(k_{\mathrm{A}_{4}}\) and \(8.25 \times 10^{-3} \mathrm{dm}^{\frac{3}{2}} / \mathrm{mol}^{\frac{1}{2}}-\mathrm{s}^{\text {in the above expression as }}\) follows:

$$ \begin{aligned} k_{\mathrm{A}} &=\frac{\left[\left(7.67 \times 10^{-3}\right)+\left(7.64 \times 10^{-3}\right)+\left(8.14 \times 10^{-3}\right)+\left(8.25 \times 10^{-3}\right)\right] \mathrm{dm}^{\frac{3}{2}} / \mathrm{mol}^{\frac{1}{2}}-\mathrm{s}}{4} \\ &=7.9 \times 10^{-3} \mathrm{dm}^{\frac{3}{2}} / \mathrm{mol}^{\frac{1}{2}}-\mathrm{s} \end{aligned} $$

Therefore, the value of \(k_{\mathrm{A}}\) is \(7.9 \times 10^{-3} \mathrm{dm}^{\frac{3}{2}} / \mathrm{mol}^{\frac{1}{2}}-\mathrm{s}\).