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Consider the reaction below that has a Keq of 8.98 X 10-2. The initial concentration of A and B is 1.68 M. Calculate the equilibrium concentration of C.
A + B
↔
2C

Respuesta :

drpelezo

Answer:

[C] = 0.4248M

Explanation:

                   A     +     B      ⇄     2C

C(i)          1.68M      1.68M         0.00

ΔC              -x            -x             +2x

C(eq)      1.68-x       1.68-x          2x

Keq = [C]²/[A][B] = (2x)²/(1.68 - x)²= 8.98 x 10⁻²

Take SqrRt of both sides => √(2x)²/(1.68 - x)² = √8.98 x 10⁻²

=> 2x/1.68 - x = 0.2895

=> 2x = 0.2895(1.68 - x)

=> 2x = 0.4863 - 0.2895x

=> 2x + 0.2895x = 0.4863

=> 2.2895x = 0.4863

=> x = 0.4863/2.2895 = 0.2124

[C] = 2x = 2(0.2124)M = 0.4248M in 'C'

Answer:

The equilibrium concentration of C is 0.4378 M

Explanation:

Step 1: Data given

Keq = 8.98 * 10^-2

The initial concentration A = 1.68 M

The initial concentration B = 1.68 M

Step 2: The balanced equation

A + B ↔ 2C

Step 3: The initial concentration

[A] = 1.68 M

[B] = 1.68 M

[C] = O M

Step 4: The concentration at the equilibrium

[A] = 1.68 -X

[B] = 1.68 - X

[C] = 2X

Step 5: Define Kc

Kc = [C]² / [A][B]

0.0898 = (2X)² / (1.68 - X)(1.68 - X)

X = 0.2189

[A] = 1.68 -0.2189 = 1.4611 M

[B] = 1.68 - 0.2189 = 1.4611 M

[C] = 2X = 0.4378 M

The equilibrium concentration of C is 0.4378 M

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