   Chapter 15, Problem 60GQ

Chapter
Section
Textbook Problem

Nitrosyl bromide, NOBr, dissociates readily at room temperature.NOBr(g) ⇄ NO(g) + ½ Br2(g)Some NOBr is placed in a flask at 25 °C and allowed to dissociate. The total pressure at equilibrium is 190 mm Hg and the compound is found to be 34% dissociated. What is the value of Kp?

Interpretation Introduction

Interpretation:

The Kp for the dissociation reaction of NOBr when the total pressure is 190mmHg has to be calculated.

Concept Introduction:

Equilibrium constant in terms of pressure[Kp]: Equilibrium constant can be expressed in terms of partial pressures in atmospheres.

aA(g)+bB(g)cC(g)+dD(g)KP=PCc×PDdPAa×PBb

The activity of solid substance will not appear in equilibrium and numerically it is considered as one.

Explanation

Given:

NOBr(g)NO(g)+12Br2(g)Totalpressure=190mmHg=0.25atm34%ofNOBrisdissociatedTemperature=25°C=298K

The partial pressures of gases are directly proportional to the number of moles, hence we can use pressures instead of moles.

NOBr(g)NO(g)+12Br2(g). This equation can be re-written as 2NOBr(g)2NO(g)+Br2(g)

2NOBr(g)2NO(g)+Br2(g)InitialpressurexatmIf34%ofNOBrisdissociated,then-0.34xisconsumedEquilibriumpressure0

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