Answer:
[tex]\boxed{1.19 \times 10^{-4} \text{ s}^{-1}}[/tex]
Explanation:
1. Determine the order of reaction
The question gives us a hint: the units of k are s⁻¹. This suggests a first order rate law.
To confirm, we plot ln(p) vs t. We should get a straight line with slope = -k.
Here are your data with the pressures converted to natural logarithms.
[tex]\begin{array}{rcc}\textbf{t/s} & \textbf{p/mmHg} &\textbf{ln(p)}\\0 & 400 & 5.99\\2000 & 316 & 5.76\\4000 & 248 & 5.51\\6000 & 196 & 5.28\\8000 & 155 & 5.04\\10000 & 122 & 4.80\\\end{array}[/tex]
We get the graph shown below.
2. Determine the rate constant
The points fit so well that we can just use the end points to determine the slope.
[tex]\text{slope} = \dfrac{y_{2} - y_{1}}{ x_{2} - x_{1} } = \dfrac{4.80 - 5.99}{10 000 - 0} = -1.19 \times 10^{-4} \text{ s}^{-1}}\\\\k = \boxed{1.19 \times 10^{-4} \text{ s}^{-1}}[/tex]
