General Chemistry
General Chemistry
4th Edition
ISBN: 9781891389603
Author: Donald A. McQuarrie, Peter A. Rock, Ethan B. Gallogly
Publisher: University Science Books
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Chapter 19, Problem 19.23P
Interpretation Introduction

Interpretation:

The total gas pressure in the vessel for the given reaction, NH4HS(s)  NH3(g) + H2S(g) has to be calculated.

Concept Introduction:

Equilibrium constant(Kp):

In gas phase reactions, partial pressure is used to write equilibrium equation than molar concentration.  Equilibrium constant (Kp) is defined as ratio of partial pressure of products to partial pressure of reactants.  Each partial pressure term is raised to a power, which is same as the coefficients in the chemical reaction

Consider the reaction where A reacts to give B.

    aAbB

    Rate of forward reaction = Rate of reverse reactionkfPAa=krPBa

On rearranging,

    PBbPAa=kfkr=Kp

Where,

    kf is the rate constant of the forward reaction.

    kr is the rate constant of the reverse reaction.

    Kp is the equilibrium constant.

Relation between Kp and Kc:

The relation between and Kp and Kc is given by the following equation.

  Kp = Kc(RT)Δngas

Where,

Kp is the equilibrium constant in terms of partial pressure

Kc is the equilibrium constant in terms of concentration

Δngas = moles of gaseous product - moles of gaseous reactant

Only moles of gaseous products and reactants is used to calculate Δngas

Expert Solution & Answer
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Explanation of Solution

Given:

Ammonium hydrogen sulphide decomposes according to the below chemical equation.

  NH4HS(s)  NH3(g) + H2S(g)

The equilibrium constant, Kc for the reaction is 1.81×104 M2 at 25°C.

Determination of total gas pressure:

First determine, Kp for the reaction using given Kc value.

Δngas  = moles of gaseous product - moles of gaseous reactant =2 - 0 =2

Then,

  Kp  = Kc(RT)Δngas = (1.81×104)[(0.0821 atm·L/mol·K)(298 K)]2 =(1.81×104)(24.4658)2 =(1.81×104)(598.58) =0.108 atm

Convert the obtained Kp value in units of bar.

  Kp = 0.108 atm = 0.109 bar

The equilibrium constant expression for the given reaction is,

  Kp = PNH3 PH2S

According to the stoichiometry of the equation, it is known that both NH3(g) and H2S(g) are present with same number of moles.

Thus, their equilibrium pressures would also be same.  Now, consider equilibrium partial pressure of each gas as ‘x’.

Using equilibrium constant expression, solve for ‘x’.

  Kp      = PNH3 PH2S 0.109 bar2 = (x)(x) 0.109 bar2 = (x)2x   = 0.109 bar2    =0.330 bar  0.33 bar

Therefore, the equilibrium partial pressures of NH3(g) and H2S(g) are,

   PNH3 = 0.33 barPH2S   = 0.33 bar

Now, sum up the obtained partial pressures and find the total gas pressure as follows,

  Total Pressure  =  PNH3 + PH2S = 0.33 bar + 0.33 bar =0.66 bar

Therefore, the total pressure in the reaction vessel at equilibrium is 0.66 bar.

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Chapter 19 Solutions

General Chemistry

Ch. 19 - Prob. 19.11PCh. 19 - Prob. 19.12PCh. 19 - Prob. 19.13PCh. 19 - Prob. 19.14PCh. 19 - Prob. 19.15PCh. 19 - Prob. 19.16PCh. 19 - Prob. 19.17PCh. 19 - Prob. 19.18PCh. 19 - Prob. 19.19PCh. 19 - Prob. 19.20PCh. 19 - Prob. 19.21PCh. 19 - Prob. 19.22PCh. 19 - Prob. 19.23PCh. 19 - Prob. 19.24PCh. 19 - Prob. 19.25PCh. 19 - Prob. 19.27PCh. 19 - Prob. 19.28PCh. 19 - Prob. 19.29PCh. 19 - Prob. 19.30PCh. 19 - Prob. 19.31PCh. 19 - Prob. 19.32PCh. 19 - Prob. 19.33PCh. 19 - Prob. 19.34PCh. 19 - Prob. 19.35PCh. 19 - Prob. 19.36PCh. 19 - Prob. 19.37PCh. 19 - Prob. 19.38PCh. 19 - Prob. 19.39PCh. 19 - Prob. 19.40PCh. 19 - Prob. 19.43PCh. 19 - Prob. 19.44PCh. 19 - Prob. 19.45PCh. 19 - Prob. 19.46PCh. 19 - Prob. 19.47PCh. 19 - Prob. 19.48PCh. 19 - Prob. 19.49PCh. 19 - Prob. 19.50PCh. 19 - Prob. 19.51PCh. 19 - Prob. 19.52PCh. 19 - Prob. 19.53PCh. 19 - Prob. 19.54PCh. 19 - Prob. 19.55PCh. 19 - Prob. 19.56PCh. 19 - Prob. 19.57PCh. 19 - Prob. 19.58PCh. 19 - Prob. 19.59PCh. 19 - Prob. 19.60PCh. 19 - Prob. 19.61PCh. 19 - Prob. 19.62PCh. 19 - Prob. 19.63PCh. 19 - Prob. 19.64PCh. 19 - Prob. 19.65PCh. 19 - Prob. 19.66PCh. 19 - Prob. 19.67PCh. 19 - Prob. 19.68PCh. 19 - Prob. 19.69PCh. 19 - Prob. 19.70PCh. 19 - Prob. 19.71PCh. 19 - Prob. 19.72PCh. 19 - Prob. 19.73PCh. 19 - Prob. 19.74PCh. 19 - Prob. 19.75PCh. 19 - Prob. 19.76PCh. 19 - Prob. 19.77PCh. 19 - Prob. 19.78PCh. 19 - Prob. 19.79PCh. 19 - Prob. 19.80PCh. 19 - Prob. 19.81PCh. 19 - Prob. 19.82PCh. 19 - Prob. 19.83PCh. 19 - Prob. 19.84PCh. 19 - Prob. 19.86PCh. 19 - Prob. 19.91PCh. 19 - Prob. 19.92PCh. 19 - Prob. 19.93PCh. 19 - Prob. 19.94P
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