Q1.) Consider the following equilibrium at 976 K for the dissociation of molecul

Question

Q1.) Consider the following equilibrium at 976 K for the dissociation of molecular iodine into atoms of iodine.

Kc = 1.51103

Suppose this reaction is initiated in a 3.9 L container with 0.072 mol I2 at 976 K. Calculate the concentrations of I2 and I at equilibrium.

(HINT: Notice that moles and volume are given - calculate the molarities first!)

(a) Determine Q at this time. (Omit units.)


(b) Determine which direction the reaction will proceed in order to reach equilibrium.

left or right   

Kc = 1.51103

Explanation / Answer

1)

initial concentration of I2,

[I2] = number of mol of I2 / volume

= 0.072 mol / 3.9 L

= 0.0185 M

I2 (g)   <—> 2I (g)

0.0185       0   (initial)

0.0185-x       2x   (at equilibrium)  

Kc = [I]^2 / [I2]

1.51*10^-3 = (2x)^2 / (0.018-x)

2.788*10^-5 - 1.51*10^-3*x = 4*x^2

4*x^2 + 1.51*10^-3*x - 2.788*10^-5 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 4.0

b = 0.00151

c = -2.788*10^-5

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 4.484*10^-4

roots are :

x = 2.458*10^-3 and x = -2.836*10^-3

since x can't be negative, the possible value of x is

x = 2.458*10^-3

[I2] = 0.0185-x = 0.185 - 2.458*10^-3 = 0.183 M

Answer: 0.183 M

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