17.86_CP - Chemistry: Molecular Nature Of Matter And Change

Question

An industrial chemist introduces $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{atm}$ of ${\mathbf{H}}_{\mathbf{2}}$ and $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{atm}$ of  into a $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{-}\mathbf{L}$ container at $\mathbf{25}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}\mathbf{C}$ and then raises the temperature to ${\mathbf{700}}^{\mathbf{°}}\mathbf{C}$, at which : ${\mathbf{K}}_{\mathbf{c}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{534}$

${\mathbf{H}}_{\mathbf{2}}\left(g\right)\mathbf{+}{\mathbf{CO}}_{\mathbf{2}}\left(g\right)\mathbf{\to }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(g\right)\mathbf{+}\mathbf{CO}\left(g\right)$

How many grams of ${\mathbf{H}}_{\mathbf{2}}$ are present at equilibrium?

Step-by-Step Solution

Answer

The amount of ${\mathrm{H}}_2$ are present at equilibrium is $0.095 \mathrm{g}$ .

Step 1: Concept Introduction

Chemical equilibrium is the state of a system in which the concentration of the reactant and the concentration of the products do not change over time and the system's attributes do not change.

Step 2: Calculation for Partial Pressures

The reaction given is –

${\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)+\mathrm{CO}\left(\mathrm{g}\right) {\mathrm{K}}_{\mathrm{c}}=0.534$

The given values are partial pressures of the reactants. Since the given equilibrium constant is in terms of concentration, we must solve for the concentration of each using the ideal gas equation. We need to use $\mathrm{T}={25}^{\mathrm{o}}\mathrm{C}=298.15 \mathrm{K}$ since the given values are the initial partial pressures.

$$\begin{gathered} \mathrm{PV}=\mathrm{nRT} \\ \frac{\mathrm{n}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{RT}} \end{gathered}$$

For Hydrogen molecule –

$$\begin{gathered} \left[{\mathrm{H}}_2\right]=\frac{2.0 \mathrm{atm}}{0.0821{\displaystyle \frac{\mathrm{atm}. \mathrm{L}}{\mathrm{mol}. \mathrm{K}}}\times 298.15 \mathrm{K}} \\ \left[{\mathrm{H}}_2\right]=0.0817 \mathrm{mol}/\mathrm{L} \end{gathered}$$

For Carbon dioxide molecule –

$$\begin{gathered} \left[{\mathrm{CO}}_2\right]=\frac{2.0 \mathrm{atm}}{0.0821{\displaystyle \frac{\mathrm{atm}. \mathrm{L}}{\mathrm{mol}. \mathrm{K}}}\times 298.15 \mathrm{K}} \\ \left[{\mathrm{CO}}_2\right]=0.0817 \mathrm{mol}/\mathrm{L} \end{gathered}$$

The reaction table is –

	${\mathrm{H}}_2 \left(\mathrm{g}\right)$	${\mathrm{CO}}_2 \left(\mathrm{g}\right)$	${\mathrm{H}}_2\mathrm{O} \left(\mathrm{g}\right)$	$\mathrm{CO} \left(\mathrm{g}\right)$
Initial	0.0817	0.0817	0	0
Change	-x	-x	+x	+x
Equilibrium	0.0817-x	0.0817-x	x	x

Step 3: Calculation for Mass

The equilibrium constant for the given reaction is –

$$\begin{gathered} {\mathrm{K}}_{\mathrm{c}}=\frac{\left[\mathrm{products}\right]}{\left[\mathrm{reactants}\right]} \\ {\mathrm{K}}_{\mathrm{c}}=\frac{\left[{\mathrm{H}}_2\mathrm{O}\right]\left[\mathrm{CO}\right]}{\left[{\mathrm{H}}_2\right]\left[{\mathrm{CO}}_2\right]} \end{gathered}$$

Substitute the values to the equilibrium constant expression and solve for x.

$\begin{array}{rcl}{\mathrm{K}}_{\mathrm{c}}& =& \frac{\left[{\mathrm{H}}_2\mathrm{O}\right]\left[\mathrm{CO}\right]}{\left[{\mathrm{H}}_2\right]\left[{\mathrm{CO}}_2\right]}\\ 0.534& =& \frac{\left(\mathrm{x}\right)\left(\mathrm{x}\right)}{\left(0.0187-\mathrm{x}\right)\left(0.0187-\mathrm{x}\right)}\\ 0.534& =& \frac{{\mathrm{x}}^2}{{\left(0.0187-\mathrm{x}\right)}^2}\\ \sqrt{0.534}& =& \frac{\mathrm{x}}{0.0187-\mathrm{x}}\\ \mathrm{x}& =& 0.0345\\ & & \end{array}$

Solve for the concentration of ${\mathrm{H}}_2$ at equilibrium.

$\begin{array}{rcl}\left[{\mathrm{H}}_2\right]& =& 0.0187-\mathrm{x}\\ & =& 0.0187-0.0345\\ \left[{\mathrm{H}}_2\right]& =& 0.0472 \mathrm{mol}/\mathrm{L}\end{array}$

Solve for the mass of $\left[{\mathrm{H}}_2\right]$ using dimensional analysis.

$\begin{array}{rcl}\mathrm{mass}& =& 0.0472 \frac{\mathrm{mol}}{\mathrm{L}}\times \frac{2.014 \mathrm{g} }{1 \mathrm{mol}}\times 1.00 \mathrm{L}\\ & =& 0.095 \mathrm{g}\\ & & \end{array}$ 

Therefore, the value of mass is obtained as 0.095 g .