Chemistry: An Atoms-Focused Approach
Chemistry: An Atoms-Focused Approach
14th Edition
ISBN: 9780393912340
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster
Publisher: W. W. Norton & Company
Question
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Chapter 14, Problem 14.82QA
Interpretation Introduction

To find:

The equilibrium partial pressures of NO2 and N2O4.

Expert Solution & Answer
Check Mark

Answer to Problem 14.82QA

Solution:

PNO2=0.276 atm  and PN2O4=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 Kp expression.

2) Formula:

Kp=PproductscoefficentsPreactantscoefficents

Here,Kp is the equilibrium partial pressure constant,Pproducts is the partial pressure of products and Preactants is the partial pressure of reactants and the coefficients are the coefficient of reactants and product from the balanced reaction.

3) Given:

i) Kp=4.0

ii) T=298 K

iii) Reaction:2NO2gN2O4(g)

iv) Initial partial pressure of NO2:  PNO2=0.900 atm

4) Calculations:

RICE table for given reaction:

Reaction 2NO2gN2O4(g)
PNO2(atm) PN2O4(atm)
Initial 0.900 0
Change -2x +x
Equilibrium (0.900-2x) x

Kp=PproductscoefficentsPreactantscoefficents

Kp=PN2O41PNO22

4.0=x0.900-2x2

4.0=x0.81-3.6x+4x2

4.0×0.81-3.6x+4x2=x

3.24-14.4x+16x2-x=0

16x2-15.4x+3.24=0(1)

This equation fits the general form of a quadratic equation:

ax2+bx+C=0

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

x=-b ± (b2-4ac)2a(2)

Here a and b are coefficients of x2, x respectively. And c is a constant term in equation 1.

Therefore, a=16, b=-15.4 and c=3.24

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

x=-(-15.4) ± -15.42-(4×16×3.242×16

x=15.4 ± (29.8)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 PNO2 and PN2O4

PNO2=0.900-2x=0.900-2×0.310658=0.900-0.621316=0.275684 atm

PN2O4=x=0.310658 atm

And,

PNO2=0.900-2x=0.900-2×0.651842=0.900-1.243684=-0.343684 atm

PN2O4=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:

PNO2=0.276 atm and PN2O4=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.

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Chapter 14 Solutions

Chemistry: An Atoms-Focused Approach

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