Chapter 14, Problem 14.82QA

Interpretation Introduction

To find:

The equilibrium partial pressures of $N{O}_2$ and $N_2{O}_4.$

Answer to Problem 14.82QA

Solution:

$P_{N{O}_2}=0.276 atm$ and $P_{N_2{O}_4}=0.311 atm$

Explanation of Solution

1) Concept:

We are using the above formula to solve this question. The values from the RICE table should be used in the $K_p$ expression.

2) Formula:

$K_p=\frac{{\left[P_{products}\right]}^{coefficents}}{{\left[P_{reactants}\right]}^{coefficents}}$

Here,$K_p$ is the equilibrium partial pressure constant,$\left[P_{products}\right]$ is the partial pressure of products and $\left[P_{reactants}\right]$ is the partial pressure of reactants and the coefficients are the coefficient of reactants and product from the balanced reaction.

3) Given:

i) $K_p=4.0$

ii) $T=298 K$

iii) $Reaction:2N{O}_{2\left(g\right)}\rightleftharpoons N_2{O}_{4\left(g\right)}$

iv) $Initial partial pressure of N{O}_2: P_{N{O}_2}=0.900 atm$

4) Calculations:

RICE table for given reaction:

Reaction	$2N{O}_{2\left(g\right)}\rightleftharpoons N_2{O}_{4\left(g\right)}$
	$P_{N{O}_2}\left(atm\right)$	$P_{N_2{O}_4}\left(atm\right)$
Initial	$0.900$	$0$
Change	$-2x$	$+x$
Equilibrium	$(0.900-2x)$	$x$

$K_p=\frac{{\left[P_{products}\right]}^{coefficents}}{{\left[P_{reactants}\right]}^{coefficents}}$

$K_p=\frac{{\left[P_{N_2{O}_4}\right]}^1}{{\left[P_{N{O}_2}\right]}^2}$

$4.0=\frac{x}{{\left(0.900-2x\right)}^2}$

$4.0=\frac{x}{0.81-3.6x+4x^2}$

$4.0\times \left(0.81-3.6x+4x^2\right)=x$

$3.24-14.4x+16x^2-x=0$

$16x^2-15.4x+3.24=0\dots \left(1\right)$

This equation fits the general form of a quadratic equation:

$a{x}^2+bx+C=0$

We need to solve this quadratic equation (1) using the following equation:

$x=\frac{-b \pm \surd (b^2-4ac)}{2a}\dots \left(2\right)$

Here $a and b$ are coefficients of $x^2, x$ respectively. And $c$ is a constant term in equation $\left(1\right).$

Therefore,$a=16, b=-15.4 and c=3.24$

Plugging these values in equation (2), we  get:

$x=\frac{-(-15.4) \pm \surd \left[{\left(-15.4\right)}^2-(4\times 16\times 3.24\right]}{2\times 16}$

$x=\frac{15.4 \pm \surd \left(29.8\right)}{32}$

Solving this equation, we can get two values of $x$ which are:$x=0.310658 and x= 0.651842$

We can use both these values to calculate the values of $P_{N{O}_2} and P_{N_2{O}_4}$

$P_{N{O}_2}=\left(0.900-2x\right)=0.900-\left(2\times 0.310658\right)=0.900-0.621316=0.275684 atm$

$P_{N_2{O}_4}=x=0.310658 atm$

And,

$P_{N{O}_2}=\left(0.900-2x\right)=0.900-\left(2\times 0.651842\right)=0.900-1.243684=-0.343684 atm$

$P_{N_2{O}_4}=x=0.651842$

We got two positive values of $x$ from the quadratic equation. But one of the values of $x$ gives a negative value of the equilibrium partial pressure of the reactant. This means that the value of $2x$ should not be greater than the initial value. Therefore, we are neglecting that value and using the one which is giving less value of equilibrium partial pressure than the initial partial pressure. Also, the valid value of $x is 0.310658$.

Therefore, the values of equilibrium partial pressures with the correct significant figures are:

$P_{N{O}_2}=0.276 atm and P_{N_2{O}_4}=0.311 atm$

Conclusion:

The partial pressure of the reactant and product at equilibrium is calculated using the initial partial pressure and the equilibrium constant.