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We have the reaction   2 A(aq) +B(aq) + 3 C(aq) ? 2 D(aq) + 2 E(aq)...

We have the reaction  
2 A(aq) +B(aq) + 3 C(aq) ? 2 D(aq) + 2 E(aq) + 3F(aq) .  
At 25oC, we start with a solution that is 0.977 M in A(aq), 0.655 M in B(aq), and 0.911 M in C(aq). There is no D(aq), E(aq), or F(aq) present. Equilibrium is established, and at equilibrium, the concentration of C(aq) is 0.677 M. What is the value of the equilibrium constant, K, for this reaction? What is the value of ?Go for this reaction? For this reaction, we have another experiment where we only start with D(aq), E(aq), and F(aq). At equilibrium, the concentrations of D(aq), E(aq), and F(aq) are, respectively, 0.211 M, 0.322 M, and 0.255 M. What is the concentration of B(aq) at equilibrium? The temperature is still 25oC.

Solutions

Expert Solution

The balanced reaction with ICE TABLE

2 A(aq) +B(aq) + 3 C(aq) = 2 D(aq) + 2 E(aq) + 3F(aq)

I 0.977 0.655 0.911   

C (-2x) (-x) (-3x) +2x +2x +3x

E (0.977-2x) (0.655-x) (0.911-3x) +2x +2x +3x

At equilibrium

concentration of C(aq) = 0.677 M

0.911-3x = 0.677

x = 0.078

Concentration of A = 0.977-2x = 0.977 - 2*0.078 = 0.821 M

Concentration of B = 0.655-x = 0.655 - 0.078 = 0.577 M

Concentration of D = concentration of E = 2*0.078 = 0.156 M

Concentration of F = 3*0.078 = 0.234 M

Equilibrium constant expression of the reaction

Kc = [F]3[E]2[D]2 / [C]3[B] [A]2

= (0.234)3(0.156)4 / (0.078)3(0.577)(0.821)2

= 0.0411

G°= - RT ln K

= - 8.314 J/mol·K x (25+273)K x ln (0.0411)

= 7907.78 J

If start with only D

At equilibrium

Kc = [F]3[E]2[D]2 / [C]3[B] [A]2

= (0.255)3(0.322)2(0.211)2 / (-3x)3(-x)(-2x)2

0.0411 = (7.654 x 10^-5) / 108x6

108x6 = 0.00186

x = 0.161

concentration of B(aq) at equilibrium = x = 0.161 M

Amanda K answered 2 years ago
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