Chapter 14, Problem 14.99QA

Interpretation Introduction

To find:

We need to find the equilibrium concentration of $NO, N_2{O}_3 and {NO}_{2 }$ gases.

Answer to Problem 14.99QA

Solution:

The equilibrium concentrations for all the species in the given reaction are:

$\left[NO\right]=0.027M$

$\left[N_2{O}_3\right]=0.098M$

$\left[N{O}_2\right]=0.277M$

Explanation of Solution

1) Concept

The system initially contains no product. This means that the reaction quotient $Q_c$ is equal to zero and thus less than K. Therefore, the reaction proceeds in the forward direction decreasing the concentration of the reactants while increasing those of the product.

Let x be the increase in the concentration of $N_2{O}_3$ as a result of the reaction. The stoichiometry of the reaction tells us that the change in the concentrations for $NO and NO$ are -x and that the change in concentration for $N_2{O}_3$ is +x. We need to find the moles of each of the species using their molar masses and then calculate the molarities.

2) Given:

An equilibrium reaction equation for formation of nitrogen trioxide is given as:

$NO\left(g\right)+N{O}_2\left(g\right)\rightleftharpoons N_2{O}_3\left(g\right)$

i) $K_p=0.535$

ii) Volume of flask is $=4L$

iii) Initial mass of $NO=15g$

iv) Initial mass of ${NO}_{2 }=69g$

3) Formula

i) $K_p= K_c{\left(RT\right)}^{\Delta n}$

Where  $\Delta n$ is number of moles of product gas minus number of gas moles of reactant gas in the balanced chemical equation.

$∆n= -1$.

ii) $K_c= \frac{{\left[C\right]}^{c }{\left[D\right]}^d}{{\left[A\right]}^a{\left[B\right]}^b}$

iii) $Molarity \left(M\right)=\frac{mol solute}{L solution}$

4) Calculations:

Initial concentration of reactants is calculated as follows,

Number of moles of $NO$ and molarity is as follows:

$mol NO=\frac{15 g}{30.006 g/mol}=0.4999 mol$

$\left[NO\right]= \frac{0.4999 mol}{4 L}=0.1249M$

Number of moles of $N{O}_2$ and molarity is calculated as:

${mol NO}_{2 }= \frac{69g}{46.005\frac{g}{mol}}=1.4998 mol$

${[NO}_{2 }]= \frac{1.4998 mol}{4 L}=0.3749M$

$K_c$ is calculated from the value of $K_p$ as follows:

$K_p= K_c{\left(RT\right)}^{\Delta n}$.

$\Delta n$. for the given reaction is -1

$0.535= K_c{(0.082\times 298)}^{-1}$

$0.535= K_c \times 0.04092$

$K_c=13.07$

Drawing the RICE table:

Reaction	$NO\left(g\right) + N{O}_2\left(g\right) \rightleftharpoons N_2{O}_3\left(g\right)$
Initial (M)	$0.1249$	$0.3749$	$0.00$
Change (M)	$-x$	$-x$	$+x$
Equilibrium (M)	$(0.1249-x)$	$(0.3749-x)$	$+x$

Inserting the equilibrium terms into the equilibrium constant expression for the reaction gives,

$K_c= \frac{\left[N_2{O}_3\right]}{\left[{NO}_{2 }\right]\left[ NO\right]}$

$13.07= \frac{x}{\left(0.1249-x\right)\times (0.3749-x)}=0.0978$

Solving for x gives $x=0.0978$ which is equilibrium concentration of $N_2{O}_3$.

The equilibrium concentration of $NO$ is calculated as follows,

$(0.1249-x)=(0.1249-0.0978)=0.027M$

The equilibrium concentration of ${NO}_{2 }$ is calculated as follows,

$0.3749-x=0.3749-0.0978=0.277M$.

Thus, equilibrium concentrations for the species $NO$ and $O_2$ are $0.277 M$ each and that for $N_2{O}_3$ is $0.098 M$

Conclusion:

Using the value of x in the terms in the Equilibrium row of the RICE table, we find the equilibrium concentration of $NO, N_2{O}_3 and {NO}_{2 }$.. These values result in $K_c$ value which is acceptably close to the given value of $13.07.$