1. For the 1st order reaction whose rate law constant, k, is 1.5x10-3s-1, t½ is
a. 7.7 minutes ***
b. 462 minutes
c. 6.5 minutes
d. 4.5 minutes
2. If the half-life of a 1st order reaction is 137 minutes, k=
a. 8.43x10-5s-1
b. 11,900 s-1
c. 5.06x10-3s-1 ***
d. 198 s-1
3. About what fraction of the original radioactive sample will be left after an hour? The half-life of the radioisotope is 15 minutes.
a. a sixteenth ***
b. an eighth
c. a quarter
d. half
1. To calculate the half-life of a 1st order reaction, we use the equation:
t½ = ln(2) / k
Given the value of k as 1.5x10-3s-1, we can substitute it into the equation:
t½ = ln(2) / 1.5x10-3s-1
Using a calculator, we can find that:
t½ = 7.7 minutes
Therefore, option a. 7.7 minutes is correct.
2. Given the half-life of a 1st order reaction as 137 minutes, we can use the same equation:
t½ = ln(2) / k
By rearranging the equation, we can solve for k:
k = ln(2) / t½
Substituting the value of t½ as 137 minutes:
k = ln(2) / 137 minutes
Using a calculator, we can find that:
k ≈ 5.06x10-3s-1
So, option c. 5.06x10-3s-1 is correct.
3. To determine the fraction of the original sample remaining after an hour for a radioactive isotope with a half-life of 15 minutes, we can calculate the number of half-lives that occur in an hour:
Number of half-lives = (60 minutes) / (15 minutes per half-life)
Number of half-lives = 4 half-lives
Since each half-life reduces the sample by half, the fraction remaining after four half-lives is:
Fraction remaining = (1/2)^4 = 1/16
Therefore, option a. a sixteenth is correct.
1. To find the half-life (t½) of a 1st order reaction, you can use the formula t½ = ln(2)/k, where k is the rate constant. In this case, the rate constant, k, is given as 1.5x10-3s-1. Plugging this value into the formula, we have t½ = ln(2)/(1.5x10-3s-1). Simplifying this expression, we find that t½ = 7.7 minutes.
2. The half-life of a 1st order reaction is given as 137 minutes. To find the rate constant (k), we can use the formula k = ln(2)/t½, where t½ is the half-life. Plugging in the given value of t½ = 137 minutes, we have k = ln(2)/(137 minutes). Simplifying this expression, we find that k = 5.06x10-3s-1.
3. The fraction of the original radioactive sample left after a certain time can be calculated using the formula N/N₀ = (1/2)^(t/t½), where N is the quantity remaining, N₀ is the original quantity, t is the time, and t½ is the half-life. In this case, the half-life is given as 15 minutes and the time is given as 1 hour (60 minutes). Plugging in these values, we have N/N₀ = (1/2)^(60/15). Simplifying this expression, we find that N/N₀ = 1/16, which is equal to a sixteenth of the original radioactive sample remaining. Therefore, the answer is option a, a sixteenth.