Q24.79P - Chemistry: Molecular Nature Of Matter And Change

Question

Oxygen-$\mathbf{16}$ is one of the most stable nuclides. The mass of a ${}^{\mathbf{\text{16}}}\mathbf{\text{O }}$atom is $\mathbf{\text{15.994915}}$amu. Calculate the binding energy 

(a) per nucleon in MeV;

(b) per atom in MeV;

 (c) per mole in kJ.

Step-by-Step Solution

Answer

The binding energies are

$$\begin{gathered} \text{a) 7.976 MeV/nucleon} \\ \text{b) 127.616 MeV/atom} \\ {\text{c) 1.231×10}}^{\text{10}}\text{ kJ/mol} \end{gathered}$$

Step 1: To calculate the binding energy(a)

- The atomic mass of ${}^{\text{16}}\text{O is 15.994915 amu}$

- Atomic number of O is  $8$ , so it has $8$ protons

- The number nucleons is $16$, hence

the number of neutrons is $\left(16-8\right)8$.

- The mass of neutron is ${\text{m}}_{\text{n}}\text{ = 1.008665amu}$- The mass of proton is ${\text{m}}_{\text{p}}\text{ = 1.007825amu}$

Now, let us calculate the change in mass

$\begin{array}{rcl}& & ∆\text{m = 15.994915amu - }\left(8\times {\text{m}}_{\text{p}}{\text{ + 8×m}}_{\text{n}}\right)\\ & & \text{ =15.994915amu - }\left(\text{8×1.007825amu + 8×1.008665amu}\right)\\ & & \text{ = - 0.137005amu}\end{array}$

(a)

The binding energy per nucleon, in  $\text{MeV}$is

$\text{ BE per nucleon = }\frac{{\displaystyle \text{0.137005amu×}\frac{\text{931.5MeV}}{\text{1amu}}}}{{\displaystyle \text{16nucleons}}}=\text{7.976MeV/ nucleon}$

Step 2: To calculate the binding energy(b)

(b)

The binding energy per atom is

$$\begin{gathered} \text{BE per atom = 7.976MeV/ nucleon×16 nucleons/atom} \\ =\text{127.616MeV/ atom} \end{gathered}$$

Step 3: To calculate the binding energy(c)

(c)

The binding energy per mole in $\text{kJ}$

$$\begin{gathered} \text{BE per mol in kJ}={\text{BE per atom×6.022×10}}^{\text{23}}\text{ atoms /moln} \\ =\text{127.616MeV/ atom×}{\mathrm{6.022\times 10}}^{\text{23}}\text{ atoms /moln} \\ ={\text{7.685×10}}^{\text{25}}\text{MeV/mol×}\frac{{\displaystyle {\text{1.602×10}}^{\text{ - 13}}\text{ J}}}{{\displaystyle \text{1MeV}}} \\ {\text{=1.231×10}}^{\text{13}}\text{ J/mol×}\frac{1 {\displaystyle \text{kJ}}}{1000 {\displaystyle \text{J}}} \\ \text{=}{\mathrm{1.231\times 10}}^{\text{10}} \text{kJ}/\mathrm{mol} \end{gathered}$$