Chapter 16, Problem 58GP

To determine

To find: The net force between the CO and HN group

Answer to Problem 58GP

  $2.4\times {10}^{-10}\text{N}$

Explanation of Solution

Given:

The C and O of CO group have charges

The H and N of HN group have charges

The given figure is shown below.

  

Formula used:

Net force between CO and HN group will be sum of forces between individual pair of charged atoms and is given by ${\mathrm{F}}_{\mathrm{n}\mathrm{e}\mathrm{t}}={\mathrm{F}}_{\mathrm{C}\mathrm{H}}+{\mathrm{F}}_{\mathrm{C}\mathrm{N}}+{\mathrm{F}}_{\mathrm{O}\mathrm{H}}+{\mathrm{F}}_{\mathrm{O}\mathrm{N}}$

  $\begin{array}{c}{\mathrm{F}}_{\mathrm{C}\mathrm{H}}= \text{Force between carbon and hydrogen }\\ =\frac{\mathrm{K}{\mathrm{q}}_{\mathrm{C}}{\mathrm{q}}_{\mathrm{H}}}{{\mathrm{r}}^2{}_{\mathrm{C}\mathrm{H}}}\end{array}$

  $\begin{array}{c}{\mathrm{F}}_{\mathrm{C}\mathrm{N}}=\text{Force between carbon and nitrogen }\\ =\frac{\mathrm{K}{\mathrm{q}}_{\mathrm{C}}{\mathrm{q}}_{\mathrm{N}}}{{\mathrm{r}}^2{}_{\mathrm{C}\mathrm{N}}}\end{array}$

  $\begin{array}{c}{\mathrm{F}}_{\mathrm{O}\mathrm{H}}= \text{Force between oxygen and hydrogen} \\ =\frac{\mathrm{K}{\mathrm{q}}_{\mathrm{O}}{\mathrm{q}}_{\mathrm{H}}}{{\mathrm{r}}^2{}_{\mathrm{O}\mathrm{H}}}\end{array}$

  $\begin{array}{c}{\mathrm{F}}_{\mathrm{O}\mathrm{N}}= \text{Force between oxygen and nitrogen }\\ =\frac{\mathrm{K}{\mathrm{q}}_{\mathrm{O}}{\mathrm{q}}_{\mathrm{N}}}{{\mathrm{r}}^2{}_{\mathrm{O}\mathrm{N}}}\end{array}$

Calculation:

  ${\mathrm{F}}_{\mathrm{n}\mathrm{e}\mathrm{t}}={\mathrm{F}}_{\mathrm{C}\mathrm{H}}+{\mathrm{F}}_{\mathrm{C}\mathrm{N}}+{\mathrm{F}}_{\mathrm{O}\mathrm{H}}+{\mathrm{F}}_{\mathrm{O}\mathrm{N}}$

Considering right direction to be positive

  $\begin{array}{l}=\left(-\frac{\mathrm{K}{\mathrm{q}}_{\mathrm{C}}{\mathrm{q}}_{\mathrm{H}}}{{\left({\mathrm{r}}_{\mathrm{C}\mathrm{H}}\right)}^2}\right)+\frac{\mathrm{K}{\mathrm{q}}_{\mathrm{C}}{\mathrm{q}}_{\mathrm{N}}}{{\left({\mathrm{r}}_{\mathrm{C}\mathrm{N}}\right)}^2}+\frac{\mathrm{K}{\mathrm{q}}_{\mathrm{O}}{\mathrm{q}}_{\mathrm{H}}}{{\left({\mathrm{r}}_{\mathrm{O}\mathrm{H}}\right)}^2}+\left(-\frac{\mathrm{K}{\mathrm{q}}_{\mathrm{O}}{\mathrm{q}}_{\mathrm{N}}}{{\left({\mathrm{r}}_{\mathrm{O}\mathrm{N}}\right)}^2}\right)\\ =\mathrm{K}\left[\frac{-0.4\mathrm{e}\times 0.2\mathrm{e}}{{\left({\mathrm{r}}_{\mathrm{C}\mathrm{H}}\right)}^2}+\frac{0.4\mathrm{e}\times 0.2\mathrm{e}}{{\left({\mathrm{r}}_{\mathrm{C}\mathrm{N}}\right)}^2}+\frac{0.4\mathrm{e}\times 0.2\mathrm{e}}{{\left({\mathrm{r}}_{\mathrm{O}\mathrm{H}}\right)}^2}-\frac{0.4\mathrm{e}\times 0.2\mathrm{e}}{{\left({\mathrm{r}}_{\mathrm{O}\mathrm{N}}\right)}^2}\right]\end{array}$

Using figure,

  $\begin{array}{c}{\mathrm{r}}_{\mathrm{C}\mathrm{H}}=0.12\text{nm}+0.28\text{nm}-0.10\text{nm}\end{array}$

  $\begin{array}{c}=0.30\text{nm}\end{array}$

  $\begin{array}{c}{\mathrm{r}}_{\mathrm{C}\mathrm{N}}=0.12\text{nm}+0.28\text{nm}\end{array}$

  $\begin{array}{c}=0.40\text{nm}\end{array}$

  $\begin{array}{c}{\mathrm{r}}_{\mathrm{O}\mathrm{H}}=0.28\text{nm}-0.10\text{nm}\end{array}$

  $\begin{array}{c}=0.18\text{nm}\end{array}$

  $\begin{array}{c}{\mathrm{r}}_{\mathrm{O}\mathrm{N}}=0.28\text{nm}\end{array}$

  $\begin{array}{c}{\mathrm{F}}_{\mathrm{n}\mathrm{e}\mathrm{t}}=\mathrm{K}\times 0.4\times 0.2\times {\mathrm{e}}^2\left[\frac{-1}{{\left(0.30\text{nm}\right)}^2}+\frac{1}{{\left(0.40\text{nm}\right)}^2}+\frac{1}{{\left(0.18\text{nm}\right)}^2}-\frac{1}{{\left(0.28\text{nm}\right)}^2}\right]\end{array}$

  $\begin{array}{c}{\mathrm{F}}_{\mathrm{n}\mathrm{e}\mathrm{t}}=\frac{9\times {10}^9{\text{Nm}}^{\text{2}}\text{/C}\times 0.8\times {(1.6\times {10}^{-19}\mathrm{C})}^2}{{(1\times {10}^{-9})}^2}\left[\frac{-1}{{(0.30)}^2}+\frac{1}{{(0.40)}^2}+\frac{1}{{(0.18)}^2}-\frac{1}{{(0.28)}^2}\right]\end{array}$

  $\begin{array}{c}{\mathrm{F}}_{\mathrm{n}\mathrm{e}\mathrm{t}}=(7.2\times 1.6\times {10}^{-10+18})\left[\frac{-1}{0.09}+\frac{1}{0.16}+\frac{1}{0.0324}-\frac{1}{0.0784}\right]\end{array}$

  $\begin{array}{c}{\mathrm{F}}_{\mathrm{n}\mathrm{e}\mathrm{t}}=(11.52\times {10}^{-8})\left[\frac{-100}{9}+\frac{100}{16}+\frac{10000}{324}-\frac{10000}{784}\right]\end{array}$

  $\begin{array}{c}{\mathrm{F}}_{\mathrm{n}\mathrm{e}\mathrm{t}}=2.44\times {10}^{-10}\text{N}\end{array}$

Conclusion:

Net force between CO and HN group is $2.44\times {10}^{-10}\text{N}$