In a chemical reaction a substance changes into another such that the ...
Given information:
A chemical reaction follows first-order kinetics, where the rate of decomposition of a chemical substance x at any instant t is proportional to the amount of unchanged substance still present.
At the end of 1 minute, half of the substance present initially has been converted.
Understanding the problem:
We need to find the time at which 99% of the substance will have changed.
Let's assume the initial amount of substance x is X.
Given that at the end of 1 minute, half of the substance has been converted, we can express this as:
X/2 = X * e^(-k*1)
Where k is the rate constant.
Solving the problem:
We can rewrite the equation as:
e^(-k) = 1/2
Taking the natural logarithm on both sides, we get:
-k = ln(1/2)
k = -ln(1/2)
Now, we need to find the time at which 99% of the substance will have changed.
Let's assume the time required for 99% change is t.
At time t, the amount of substance x remaining will be X * e^(-k*t).
We can express this as:
X * e^(-k*t) = 0.01X
Dividing both sides by X, we get:
e^(-k*t) = 0.01
Taking the natural logarithm on both sides, we get:
-k*t = ln(0.01)
Substituting the value of k, we have:
-(-ln(1/2))*t = ln(0.01)
t = ln(0.01)/ln(1/2)
t = -ln(0.01)/ln(2)
t ≈ 6.64
Answer:
The time at which 99% of the substance will have changed is approximately log2 100 minutes (option D).