26. the half - life of sulfur - 35 is 87.32 days. sulfur - 35 is used in radiation therapy to prevent…

Problem 26

a.

Explanation:

Step1: Recall half - life formula for first - order reaction

The formula for the half - life ($t_{1/2}$) of a first - order reaction is $t_{1/2}=\frac{\ln 2}{k}$, where $k$ is the rate constant. We can solve for $k$. $k = \frac{\ln 2}{t_{1/2}}$

Step2: Substitute the given half - life value

Given $t_{1/2}=87.32$ days, so $k=\frac{\ln 2}{87.32}\text{ days}^{-1}\approx\frac{0.693}{87.32}\text{ days}^{-1}\approx 7.94\times 10^{-3}\text{ days}^{-1}$

Answer:

$k\approx7.94\times 10^{-3}\text{ days}^{-1}$

b.

Explanation:

Step1: Determine the fraction remaining

If 60.0% decays, then the fraction remaining $N/N_0=1 - 0.600 = 0.400$. For a first - order reaction, the integrated rate law is $\ln\left(\frac{N}{N_0}\right)=-kt$. We can solve for $t$. $t=-\frac{\ln\left(\frac{N}{N_0}\right)}{k}$

Step2: Substitute the values of $k$ and $\frac{N}{N_0}$

We know $k = 7.94\times 10^{-3}\text{ days}^{-1}$ and $\frac{N}{N_0}=0.400$. So $t=-\frac{\ln(0.400)}{7.94\times 10^{-3}\text{ days}^{-1}}=\frac{-(- 0.916)}{7.94\times 10^{-3}\text{ days}^{-1}}\approx115\text{ days}$

Answer:

$t\approx115\text{ days}$

c.

Explanation:

Step1: Use the integrated rate law

The integrated rate law for a first - order reaction is $\ln\left(\frac{N}{N_0}\right)=-kt$. First, find $k = 7.94\times 10^{-3}\text{ days}^{-1}$ and $t = 25.6$ days. Then $\ln\left(\frac{N}{N_0}\right)=-7.94\times 10^{-3}\text{ days}^{-1}\times25.6\text{ days}\approx - 0.203$

Step2: Solve for $N$

Exponentiating both sides, $\frac{N}{N_0}=e^{- 0.203}\approx0.816$. Given $N_0 = 500.0$ g, then $N=0.816\times500.0\text{ g}=408\text{ g}$

Answer:

$N = 408\text{ g}$

Problem 27

a.

Explanation:

Step1: Recall half - life formula for first - order reaction

Using $t_{1/2}=\frac{\ln 2}{k}$, we solve for $k$. So $k=\frac{\ln 2}{t_{1/2}}$

Step2: Substitute the given half - life value

Given $t_{1/2}=15.00$ hours, then $k=\frac{\ln 2}{15.00}\text{ hours}^{-1}\approx\frac{0.693}{15.00}\text{ hours}^{-1}=4.62\times 10^{-2}\text{ hours}^{-1}$

Answer:

$k = 4.62\times 10^{-2}\text{ hours}^{-1}$

b.

Explanation:

Step1: Determine the fraction remaining

The fraction remaining $N/N_0 = 0.220$. Using the integrated rate law $\ln\left(\frac{N}{N_0}\right)=-kt$, we solve for $t$. So $t=-\frac{\ln\left(\frac{N}{N_0}\right)}{k}$

Step2: Substitute the values of $k$ and $\frac{N}{N_0}$

We know $k = 4.62\times 10^{-2}\text{ hours}^{-1}$ and $\frac{N}{N_0}=0.220$. Then $t=-\frac{\ln(0.220)}{4.62\times 10^{-2}\text{ hours}^{-1}}=\frac{-(-1.514)}{4.62\times 10^{-2}\text{ hours}^{-1}}\approx32.8\text{ hours}$

Answer:

$t\approx32.8\text{ hours}$

c.

Explanation:

Step1: Use the integrated rate law

The integrated rate law is $\ln\left(\frac{N}{N_0}\right)=-kt$. Here, $k = 4.62\times 10^{-2}\text{ hours}^{-1}$ and $t = 48.00$ hours. So $\ln\left(\frac{N}{N_0}\right)=-4.62\times 10^{-2}\text{ hours}^{-1}\times48.00\text{ hours}\approx - 2.22$

Step2: Solve for $N$

Exponentiating both sides, $\frac{N}{N_0}=e^{-2.22}\approx0.110$. Given $N_0 = 45.0$ g, then $N=0.110\times45.0\text{ g}=4.95\text{ g}$

Answer:

$N = 4.95\text{ g}$

d.

Explanation:

Step1: Write the beta - decay equation

In beta decay, a neutron in the nucleus is converted into a proton and an electron (beta particle). The atomic number increases by 1 and the mass number remains the same. For sodium - 24 (${11}^{24}Na$), the beta - decay equation is ${11}^{24}Na\rightarrow_{12}^{24}Mg + _{- 1}^{0}e$

Answer:

${11}^{24}Na\rightarrow{12}^{24}Mg + _{- 1}^{0}e$