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. Part A Kc= 1.8 Express your answer using two significant figures. Enter your answers numerically separated by commas. [A][ A ], [B], [C] = nothing M

Answer 1

A (g) $\rightleftharpoons$ B(g) + C(g)

ICE table is

	[A]	[B]	[C]
Initial	1.0	0	0
Change	-x	+x	+x
Equilibrium	(1.0-x)	x	X

K_c =

[B][C] AB

Or, 1.8 =

X T (1 )

Or, 1.8(1-x) = x²

Or, 1.8 - 1.8x = x².

Or, x² +1.8x -1.8 = 0

Using solve for quadratic equation.

X =

bt vb-4ac 2a

a = 1 , b = 1.8 , c = -1.8

X =

1.824 1* (-1.8) 1.8 2 1

Or, x =

1.8 t3.23 2 1

Or, x = 0.715 or, X = - 2.515

Final molarity cannot be negative.

Hence, x = 0.715 M.

So, [A] = (1.0 - 0.715) = 0.285 M = 0.28 M ( 2 significant figure)

[B] = 0.715 M = 0.72 M ( 2 significant figure)

[C] = 0.715 M = 0.72 M ( 2 significant figure)

[Putiing the value un expression of K_c

K_c = = 1.8

0.715 x 0.715 0.285

Hence answer is correct.]