# Answer to Question #145954 in Inorganic Chemistry for Diamond

Question #145954
A mixture of 0.300 mol H2 and 0.300 mol N2 was placed in a 1.00L stainless-steel flask at 430°C. The equilibrium constant Kc for the following reaction is 0.65 at 395°C. The gas constant R=0.08206 L.Atm/mol.K
N2(g) + 3H2(g) = 2NH3(g) (PS: please take the = sign to mean a reversible reaction).
a) what's the value of Kp for this reaction?
b) what's the value of the equilibrium constant Kc for
2NH3(g) = N2(g) + 3H2(g)
c) What's the value of the equilibrium constant for Kc for
1/2N2(g) + 3/2H2(g) = NH3(g)
d) What are the values of Kp for the reactions described in (b) and (c)
Thanks!!!!!!!!
1
2020-11-23T06:57:42-0500

A)To find Kp we will use the formula:

Kp=Kc (RT)^dn where dn is equal to the sum of the coefficient of the product minus that of the reactant which is 2-(3+1)=-2

Kp=0.65(0.08206×668)^-2

It can be written as

Kp=0.65/(0.08206×668)^2

Kp=0.0002

B)Since the reaction was reversed,then we can get the Kc of the reaction by taking the inverse of the Kc of the original reaction

Doing that we have Kc=1/0.0002

=5000

C)To find this ,we need to get the real molar concentrations of the reactants

The stoichiometry of the reaction is 3mol of H2 reacting with 1 mol of N2 to yield 2 mol of NH3.

Let x be the depletion in concentration (mol/L) of N2 and H2 at equilib￾rium. It follows that the equilibrium concentration of NH3 must be 2x. We summa￾rize the changes in concentration as follows:

N2 + 3H2---------2NH3

Initial 0.3 0.3 0

Change X 3x 2x

Equilibrium 0.3-x 0.3-3x 2x

Input the equilibrium values into the original reaction which is [NH3]^2/[N2][H2]^3 and equating it our first Kc,we have;

0.65=(2x)^2/(0.3-x)(0.3-3x)

Solving and collecting like terms, we have the quadratic equation;

34x^2+13x-1=0

Solving the quadratic equation, we jave 2 values of x which are 0.065 and -0.448

But the second value of x is physically impossible because the amounts of N2 and H2

reacted would be more than those originally present.So the value of x is 0.065

Now to calculate the molar concentrations

N2=0.3-x=0.3-0.065=0.235M

H2=0.3-3x=0.3-3(0.065)=0.105M

NH3=2x=2×0.065=0.13M

Now to get the Kc for the third reaction which is

1/2N2(g) + 3/2H2(g) = NH3(g)

Kc= [NH3]/[N2]^1/2[H2]^3/2

Kc=[0.13]/[0.235]^1/2[H2]^3/2

Kc=7.9

D)kp for the reaction in *B* will be

Kp=5000×(668×0.08206)^2

Kp=1.5×10^7

Kp for the reaction in *C* will be

dn for the reaction is -1

Kp=7.9×(0.08206 ×668)^-1

Kp=7.9/(0.08206×668)

Kp=0.144

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