The reaction SO2 Cl2 gives SO2 +Cl2 is a first-order reaction with a half-life of 3.5 * 10^4 s at 320 K. What percentage of SO2Cl2 would be decomposed on heating at 320 K for 920 minutes?

To determine the percentage of SO2Cl2 that would decompose when heated at 320 K for 920 minutes, we first need to understand the relationship between time, half-life, and the amount of substance remaining in a first-order reaction. Given that the half-life of the reaction is 3.5 x 10^4 seconds, we can use this information to calculate how many half-lives fit into the total time of 920 minutes.

Calculating Total Time in Seconds

First, let's convert 920 minutes into seconds:

• 920 minutes x 60 seconds/minute = 55,200 seconds

Finding the Number of Half-Lives

Next, we can find out how many half-lives fit into 55,200 seconds:

• Number of half-lives = Total time / Half-life
• Number of half-lives = 55,200 seconds / (3.5 x 10^4 seconds) ≈ 1.577

Calculating Remaining Amount of SO2Cl2

In a first-order reaction, the amount of substance remaining after a certain number of half-lives can be calculated using the formula:

• Remaining amount = (1/2)^(number of half-lives)

Substituting the number of half-lives we calculated:

• Remaining amount = (1/2)^(1.577) ≈ 0.396

Determining the Percentage Decomposed

To find the percentage of SO2Cl2 that has decomposed, we can subtract the remaining amount from 1 (or 100%):

• Percentage decomposed = (1 - Remaining amount) x 100
• Percentage decomposed = (1 - 0.396) x 100 ≈ 60.4%

Final Result

Thus, approximately 60.4% of SO2Cl2 would be decomposed after heating at 320 K for 920 minutes. This calculation illustrates how the concept of half-life is crucial in understanding the kinetics of chemical reactions, especially in predicting how much reactant remains after a given period.