Question

Consider the following reaction: A(g)⇌B(g)+C(g) Find the equilibrium concentrations of A, B, and C for each of the following different values of Kc. Assume that the initial concentration of A in each case is 1.0 M and that the reaction mixture initially contains no products. Make any appropriate simplifying assumptions.

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Part A

Kc= 1.8

Express your answer using two significant figures. Enter your answers numerically separated by commas.

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Part B

Kc= 1.0×10⁻²

Express your answer using two significant figures. Enter your answers numerically separated by commas.

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Part C

Kc= 2.0×10⁻⁵

Express your answer using two significant figures. Enter your answers numerically separated by commas.

Answer 1

Given, The equilibrium reaction,

A(g) B(g) + C(g)

[A] = 1.0 M

Part A) Given,

K_c = 1.8

Now, drawing an ICE chart,

	A(g)	B(g)	C(g)
I(M)	1.0	0	0
C(M)	-x	+x	+x
E(M)	1.0-x	x	x

Now, the K_c expression is,

K_c = [B][C] / [A]

1.8 = [x][x] /[1.0-x]

1.8 - 1.8x -x² = 0

x² + 1.8x -1.8 = 0

Solving the quadratic equation,

x = 0.7155

Now, from the ICE chart,

[A]_eq = [1.0-x] = 0.28 M [2 S.F]

[B]_eq = [x] = 0.72 M [2 S.F]

[C]_eq = [x] = 0.72 M [2 S.F]

Part B) Given,

K_c = 1.0 x 10^-2

Now, drawing an ICE chart,

	A(g)	B(g)	C(g)
I(M)	1.0	0	0
C(M)	-x	+x	+x
E(M)	1.0-x	x	x

Now, the K_c expression is,

K_c = [B][C] / [A]

1.0 x 10^-2 = [x][x] /[1.0-x]

0.01 - 0.01x -x² = 0

x² + 0.01x -0.01 = 0

Solving the quadratic equation,

x = 0.095

Now, from the ICE chart,

[A]_eq = [1.0-x] = 0.905 M Or 0.90 M [2S.F]

[B]_eq = [x] = 0.095 M [2 S.F]

[C]_eq = [x] = 0.095 M [2 S.F]

Part C) Given,

K_c = 2.0 x 10^-5

Now, drawing an ICE chart,

	A(g)	B(g)	C(g)
I(M)	1.0	0	0
C(M)	-x	+x	+x
E(M)	1.0-x	x	x

Now, the K_c expression is,

K_c = [B][C] / [A]

2.0 x 10^-5 = [x][x] /[1.0-x]

2.0 x 10^-5 = [x]² /[1.0] -----Here, [1.0-x] 1.0, since x <<< 1.0

x = 0.00447

Now, from the ICE chart,

[A]_eq = [1.0-x] = 1.0 [2S.F]

[B]_eq = [x] = 0.0045 M [2 S.F]

[C]_eq = [x] = 0.0045 M [2 S.F]