(b) In the reduction of nitric oxide, 50% of the reaction was completed in 108 seconds when the initial pressure was 336 mm Hg, and in 147 seconds when the initial pressure was 288 mm Hg. Find the order of the reaction.

(5 marks)



Answer :

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.