Question

The general form of the rate law for this reaction is \[ \text { Rate }=k[\mathrm{NO}]^{x}\left[\mathrm{Cl}_{2}\right]^{y} \] We can determine the values of \( x \) and \( y \) by comparing the rates from the various experiments. To determine the value of \( y \), we use the results from experiments 1 and 2 , in which only \( \left[\mathrm{Cl}_{2}\right] \) changes. In these experiments, \( [\mathrm{NO}] \) is held constant and \( \left[\mathrm{Cl}_{2}\right] \) is doubled. The rate also doubled. Thus the reaction is first order with respect to \( \left[\mathrm{Cl}_{2}\right] \). Or mathematically, \[ \begin{array}{l} \frac{0.36}{0.18}=\frac{k(0.10)^{x}(0.20)^{y}}{k(0.10)^{x}(0.10)^{y}}=\frac{(0.20)^{y}}{(0.10)^{y}} \\ 2.0=(2.0)^{y} \\ y=1 \end{array} \] To determine the value of \( x \), we use the results from experiments 2 and 3 , in which only \( [\mathrm{NO}] \) changes. Here, as the \( \mathrm{NO} \) concentration doubles \( \left(\mathrm{Cl}_{2}\right. \) concentration is constant), the rate increases by a factor of four. Thus the reaction is second order with respect to NO. Or mathematically: \[ \begin{array}{l} \frac{1.44}{0.36}=\frac{k(0.20)^{x}(0.20)^{y}}{k(0.10)^{x}(0.20)^{y}}=\frac{k(0.20)^{x}}{k(0.10)^{x}} \\ 4.0=(2.0)^{x} \\ x=2 \end{array} \] The rate law can now be written: \[ \text { Rate }=k[\mathrm{NO}]^{2}\left[\mathrm{Cl}_{2}\right] \] b What is the value of the rate constant? Rate constant \( = \) \( \square \) \[ \frac{\mathrm{L}^{2}}{\mathrm{~mol}^{2} \cdot \min } \]

          The general form of the rate law for this reaction is
\[
\text { Rate }=k[\mathrm{NO}]^{x}\left[\mathrm{Cl}_{2}\right]^{y}
\]

We can determine the values of \( x \) and \( y \) by comparing the rates from the various experiments. To determine the value of \( y \), we use the results from experiments 1 and 2 , in which only \( \left[\mathrm{Cl}_{2}\right] \) changes.

In these experiments, \( [\mathrm{NO}] \) is held constant and \( \left[\mathrm{Cl}_{2}\right] \) is doubled. The rate also doubled. Thus the reaction is first order with respect to \( \left[\mathrm{Cl}_{2}\right] \). Or mathematically,
\[
\begin{array}{l}
\frac{0.36}{0.18}=\frac{k(0.10)^{x}(0.20)^{y}}{k(0.10)^{x}(0.10)^{y}}=\frac{(0.20)^{y}}{(0.10)^{y}} \\
2.0=(2.0)^{y} \\
y=1
\end{array}
\]

To determine the value of \( x \), we use the results from experiments 2 and 3 , in which only \( [\mathrm{NO}] \) changes. Here, as the \( \mathrm{NO} \) concentration doubles \( \left(\mathrm{Cl}_{2}\right. \) concentration is constant), the rate increases by a factor of four. Thus the reaction is second order with respect to NO. Or mathematically:
\[
\begin{array}{l}
\frac{1.44}{0.36}=\frac{k(0.20)^{x}(0.20)^{y}}{k(0.10)^{x}(0.20)^{y}}=\frac{k(0.20)^{x}}{k(0.10)^{x}} \\
4.0=(2.0)^{x} \\
x=2
\end{array}
\]

The rate law can now be written:
\[
\text { Rate }=k[\mathrm{NO}]^{2}\left[\mathrm{Cl}_{2}\right]
\]
b What is the value of the rate constant?
Rate constant \( = \) \( \square \)
\[
\frac{\mathrm{L}^{2}}{\mathrm{~mol}^{2} \cdot \min }
\]

The general form of the rate law for this reaction is

Rate =k[NO]^x[Cl2]^y

We can determine the values of x and y by comparing the rates from the various experiments. To determine the value of y, we use the results from experiments 1 and 2 , in which only [Cl2] changes.

In these experiments, [NO] is held constant and [Cl2] is doubled. The rate also doubled. Thus the reaction is first order with respect to [Cl2]. Or mathematically,

(0.36)/(0.18)=(k(0.10)^x(0.20)^y)/(k(0.10)^x(0.10)^y)=((0.20)^y)/((0.10)^y)

2.0=(2.0)^y

y=1

To determine the value of x, we use the results from experiments 2 and 3 , in which only [NO] changes. Here, as the NO concentration doubles (Cl2. concentration is constant), the rate increases by a factor of four. Thus the reaction is second order with respect to NO. Or mathematically:

(1.44)/(0.36)=(k(0.20)^x(0.20)^y)/(k(0.10)^x(0.20)^y)=(k(0.20)^x)/(k(0.10)^x)

4.0=(2.0)^x

x=2

The rate law can now be written:

Rate =k[NO]^2[Cl2]

b What is the value of the rate constant?
Rate constant = □

(L^2)/( mol^2·min)

Added by Karen F.

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