10% of a sample reacts in 25 s. If the reaction follows first order kinetics, how long would it take for half the sample to react?
ln(No/N) = kt
If we start with an arbitrary number, say 100, and 10% is used, that would leave 90
ln(100/90) = k(25sec)
solve for k. You want the half life; therefore, substitute the k you find in the same equation but this time start with 100 and N = 50 (half is gone). Solve for t and that will be t1/2
To find out how long it would take for half the sample to react, we need to calculate the half-life, which is the time it takes for the quantity of a substance to reduce by half.
In this case, we are given that 10% of the sample reacts in 25 seconds, which implies that 90% of the sample remains after 25 seconds. This suggests that the reaction is an exponential decay process.
The half-life can be determined using the equation for first-order kinetics:
t(1/2) = (ln 2) / k
In this formula:
- t(1/2) represents the half-life
- ln 2 is the natural logarithm of 2, which is approximately 0.693 (ln 2 ≈ 0.693)
- k represents the rate constant for the reaction
Since we are not given the rate constant directly, we can calculate it using the given data.
Step 1: Calculate the rate constant (k)
We know that 10% of the sample reacts in 25 seconds, so the remaining 90% reacts in the remaining time (t).
ln(0.90) = -k * 25
Solving this equation, we can find the value of k:
k ≈ - (ln(0.90))/25
Step 2: Calculate the half-life (t(1/2))
Now that we have the value for k, we can substitute it into the half-life equation:
t(1/2) = (ln 2) / k
Substituting the value of k:
t(1/2) ≈ (ln 2) / [-(ln(0.90))/25]
Simplifying this equation will give us the time required for half the sample to react.