At particular concentration, the half life of the reaction is 100 minu...
Solution
Given
Concentration = C
Half-life = 100 min
Concentration doubled = 2C
New half-life = 25 min
Formula
The formula for calculating the half-life of a reaction is as follows:
t
1/2 = (0.693/k)
where,
t
1/2 = Half-life of the reaction
k = Rate constant of the reaction
Explanation
Let's assume that the order of the reaction is 'n'. Therefore, the rate equation for the reaction can be written as:
Rate = k[C]
nSubstituting the given values, we get the following equations:
For the first case,
t
1/2 = 100 min
Rate = k[C]
n Substituting t
1/2 in the formula, we get:
100 = (0.693/k)[C]
-n Taking the natural logarithm on both sides, we get:
ln(100) = ln(0.693/k) - nln[C]
ln(100) - ln(0.693/k) = -nln[C]
ln(100/0.693) = nln[C]
n = ln(100/0.693)/ln[C] - - - - (1)
For the second case,
t
1/2 = 25 min
Rate = k[2C]
n Substituting t
1/2 in the formula, we get:
25 = (0.693/k)[2C]
-n Simplifying the equation, we get:
ln(2) + ln(25) = ln(0.693/k) - nln[2C]
ln(50/0.693) = nln[2C] - ln(0.693/k)
n = ln(50/0.693)/(ln[2C]) - - - - (2)
Dividing equation (2) by equation (1), we get:
n/n = ln(50/0.693)ln[C]/ln(100/0.693)ln[2C]
Simplifying the equation, we get:
ln(2) = ln[C]/ln[2C]
2ln(2) = ln[C] - ln[2C]
ln[C] = 2ln(2) + ln[2C]
ln[C] = ln(2
2) + ln(2
1C)
ln[C] = ln(4) + ln(C)
ln[C] - ln(C) = ln(4)