Question

Question: Dry air containing a small amount of $\mathbf{CO}$was passed through a tube containing a catalyst for the oxidation of $\mathbf{CO}$to${\mathbf{CO}}_{\mathbf{2}}$ . Because of the heat evolved in this oxidation, the temperature of the air increased by$\mathbf{3}\mathbf{.}\mathbf{2}\mathit{K}$ .

Calculate the weight percentage of $\mathbf{CO}$in the sample of air. Assume that the specific heat at constant pressure for air is $\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{}\mathbf{J}\mathbf{}{\mathbf{K}}^{\mathbf{-}\mathbf{1}}\mathbf{}{\mathbf{g}}^{\mathbf{-}\mathbf{1}}$.

Short answer

The weight percentage of $\mathrm{CO}$ in the sample of air is $0.032\%$.

Step-by-step solution

Given data

Dry air containing a small amount of$\mathrm{CO}$was passed through a tube where it oxidized to${\mathrm{CO}}_2$.

Because of the heat evolved in this reaction the temperature raises$32K$.

The change in temperature$\Delta T=32\mathrm{K}$.

The value of heat capacity $c_p=1.01{\mathrm{K}}^{-1}{\mathrm{g}}^{-1}.$

Concept of change in enthalpy of reaction

Enthalpy change is the standard enthalpy of formation, which has been determined for a vast number of substances.

The reactants undergo chemical changes and combine to produce products in any general chemical reaction.

It can be represented by the following equation:$\mathbf{Reactant}\to \mathbf{Product}$

For any such reaction, the change in enthalpy is represented as$\mathit{\Delta }\mathit{H}$ and is termed the reaction enthalpy.

The reaction enthalpy is calculated as $\mathit{\Delta }\mathit{H}\mathbf{=}\mathit{\Delta }{\mathit{H}}_{\mathbf{1}}\mathbf{+}\mathit{\Delta }{\mathit{H}}_{\mathbf{2}}$.

Where,$\Delta H_1$ is change in enthalpy of reactant and $\Delta H_2$ is change in enthalpy of product.

Calculation of enthalpy change

The overall change in enthalpy for a reaction is,

$\Delta H=\sum \Delta H\left(\mathrm{products}\right)-\sum \Delta H\left(\mathrm{reactants}\right)$.

The standard enthalpies:

$\begin{array}{c}\Delta H_f\left({\mathrm{CO}}_2,g\right)=-393.51\mathrm{kJ}{\mathrm{mol}}^{-1}\\ \Delta H_f(\mathrm{CO},s)=-110.52{\mathrm{kmol}}^{-1}\end{array}$

In our case the equation would look like this:

$\begin{array}{c}\Delta H=2\Delta H_f\left({\mathrm{CO}}_2,g\right)-2\Delta H_f(\mathrm{CO},s)\\ \Delta H=2\times (-393.51)-2\times (-110.52)\\ \Delta H=-565.98\mathrm{kJ}\end{array}$

For above process the relation of heat and enthalpy change is given below.

$\begin{array}{l}q=-\Delta H\\ q=565.98\mathrm{kJ}\end{array}$

Calculation of mass of  CO

Because the reaction of oxidation releases heat.

The equation for the heat is, $q=m{c}_s\Delta T$.

This means that the mass of the air sample will be calculated from the formula.

$\begin{array}{c}m=\frac{q}{c_s\Delta T}\\ m=\frac{565.98\times {10}^3}{1.01\times 3.2}\\ m_{\mathrm{sample}}=175.12\mathrm{kg}\end{array}$

Calculating the mass of CO.

$M\left(\mathrm{CO}\right)=28.01\mathrm{g}{\mathrm{mol}}^{-1}$

Using the reaction's stoichiometry, $n\left(\mathrm{CO}\right)=2\mathrm{mol}$

$\begin{array}{l}m=n\left(\mathrm{CO}\right)\times M\left(\mathrm{CO}\right)\\ m\left(\mathrm{CO}\right)=2\times 28.01\\ m\left(\mathrm{CO}\right)=56.02\mathrm{g}\end{array}$

Calculation of mass percentage of compound

The formula for the mass percentage of a compound in a sample is

$w\left(\mathrm{compound}\right)=\frac{m\left(\mathrm{compound}\right)}{m\left(sample\right)}\times 100\%$

Calculation of the mass percentage.

$\begin{array}{c}m\left(\mathrm{CO}\right)=56.02\mathrm{g}\\ m\left(\mathrm{sample}\right)=175.12\times {10}^3\mathrm{g}\\ w\left(\mathrm{CO}\right)=\frac{56.02}{175\times {10}^3}\times 100\%\\ w\left(\mathrm{CO}\right)=0.032\%\end{array}$