1. The first order rate constant for the reaction, SO2Cl2(g) → SO2(g) + Cl2(g), is 2.20 x 10-5 s-1 at 593 K. What percent of a sample of SO2Cl2 would be decomposed by heating at 593 K for (a) 1 hr, (b) 3 hr. How long will it take for half of the SO2Cl2 to decompose?
ln(No/N) = kt
(c) part.
Let No = 100 to start.
then N = 50 at end.
ln(100/50) = 2.2E-5*t
0.693 = 2.2E-5*t and
t = about 30,000 hours but that's just a close estimate. You should get a better answer. Post your work if you have problems.
Ka re nahi aa raha hai. Ka karta hai.
To determine the percent of a sample of SO2Cl2 decomposed over a given time, you can use the first-order rate constant and the formula for exponential decay:
(a) To find the percent of the sample decomposed over 1 hour:
First, convert 1 hour to seconds: 1 hour * 60 min/hour * 60 sec/min = 3600 seconds
Next, use the formula for exponential decay:
Percent decomposed = (1 - e^(-kt)) * 100
where k is the rate constant and t is the time in seconds.
Plug in the given values:
Percent decomposed = (1 - e^((-2.20 x 10^-5 s^-1) * (3600 s))) * 100
Now you can calculate the result using a scientific calculator or programming language that supports exponential calculations.
(b) Similarly, to find the percent of the sample decomposed over 3 hours:
Convert 3 hours to seconds: 3 hours * 60 min/hour * 60 sec/min = 10800 seconds
Percent decomposed = (1 - e^(-kt)) * 100
where k is the rate constant and t is the time in seconds.
Plug in the given values:
Percent decomposed = (1 - e^((-2.20 x 10^-5 s^-1) * (10800 s))) * 100
To find the time it takes for half of the SO2Cl2 to decompose, you can use the formula for the half-life of a first-order reaction:
t(1/2) = ln(2) / k
where t(1/2) is the half-life, ln(2) is the natural logarithm of 2, and k is the rate constant.
Plug in the given rate constant:
t(1/2) = ln(2) / (2.20 x 10^-5 s^-1)
Now you can calculate the result using a scientific calculator or programming language that supports natural logarithm calculations.