Question

a)For a first order decay of [A], if 188 mg remains of an initial sample of 1.3737 g after 516 min, what is the half life (in minutes)?

b)The decay of carbon-14 is first order with a half life is 5576 years. How much of 1.4376 g sample would remain after 9190 years?

c)The decay of cesium-135 is first order with a half life is 3.0 million years. How long (in years) would it take for a 0.9798 g sample to be reduced to 0.2518 g?

Answer 1

a)

Step i) first calculate the rate constant using integrated formula for first order reaction:

ln[A₀] - ln[A_t] = kt ...(1)

[A₀] = Initial concentration = 1.3737 g

[A_t] = concentration after time t = 188 mg = 0.188 g

t = 516 min

Substituting these values in equation (i);

ln(1.3737) - ln(0.188) = 516 × k

1.9888 = 516 × k

K = 0.003854 min^-1

Step ii) Now calculate half life using following formula;

t_1/2 = 0.693/k = 0.693/0.003854

t_1/2 = 179.8 min.

b) step i) first calculate rate constant using following formula:

t_1/2 = 0.693/k

t_1/2 = 5576 years

k = 0.693/5576

k = 1.2428×10^-4 year^-1

step ii) Now calculate sample remained after 9190 years using integrated for first order reaction:

ln[A₀] - ln[A_t] =kt

Here t=9190 years

[A₀] = 1.4376

ln(1.4376)-ln[A_t] = 1.2428×10^-4 × 9190

[A_t]=0.4587 g

c) step i) use following equations to calculate rate constant:

_t1/2 = 0.693/k

t_1/2 = 3 million years = 3 × 10⁶ year

k = 0.693/(3×10⁶)

k = 2.31 × 10^-5 year^-1

^{step ii) now claculate time required for given decay using integrated formula for first order reaction:}

ln(0.9798)-ln(0.2518)=2.31×10^-5×t

t = 5.7 billion years

Here [A