To determine the order of the reaction, we need to use the times for the reaction to reach 50% completion at different initial pressures.
Given data:
- [tex]\( t_1 = 108 \)[/tex] seconds for 50% completion at [tex]\( P_1 = 336 \)[/tex] mm Hg
- [tex]\( t_2 = 147 \)[/tex] seconds for 50% completion at [tex]\( P_2 = 288 \)[/tex] mm Hg
For a reaction of order [tex]\( n \)[/tex], the relationship between half-life and initial pressure is given by:
[tex]\[ \left(\frac{t_1}{t_2}\right) = \left(\frac{P_2}{P_1}\right)^{n-1} \][/tex]
Rearranging the equation to solve for [tex]\( n \)[/tex]:
[tex]\[ n - 1 = \frac{\log(t_1/t_2)}{\log(P_2/P_1)} \][/tex]
[tex]\[ n = 1 + \frac{\log(t_1/t_2)}{\log(P_2/P_1)} \][/tex]
### Step-by-step solution:
1. Compute the ratio of times:
[tex]\[ \frac{t_1}{t_2} = \frac{108}{147} \approx 0.7347 \][/tex]
2. Compute the ratio of pressures:
[tex]\[ \frac{P_2}{P_1} = \frac{288}{336} \approx 0.8571 \][/tex]
3. Calculate the logarithm of the time ratio:
[tex]\[ \log(t_1/t_2) = \log(0.7347) \approx -0.1332 \][/tex]
4. Calculate the logarithm of the pressure ratio:
[tex]\[ \log(P_2/P_1) = \log(0.8571) \approx -0.0674 \][/tex]
5. Compute the order of the reaction [tex]\( n \)[/tex]:
[tex]\[ n = 1 + \frac{\log(t_1/t_2)}{\log(P_2/P_1)} \][/tex]
[tex]\[ n = 1 + \frac{-0.1332}{-0.0674} \][/tex]
[tex]\[ n = 1 + 1.9758 \][/tex]
[tex]\[ n \approx 2.976 \][/tex]
Given the result from our calculations, the order of the reaction is remarkably close to an integer value.
Therefore, the order of the reaction is 3.
