21.Be sure to answer all parts. The dissociation of molecular iodine into iodine

Question

21.Be sure to answer all parts.

The dissociation of molecular iodine into iodine atoms is represented as
I2(g) 2I(g)
At 1000 K, the equilibrium constant Kc for the reaction is 3.80 ×105. Suppose you start with 0.0461 mol of I2 in a 2.27L flask at 1000 K. What are the concentrations of the gases at equilibrium?

What is the equilibrium concentration of I2?

What is the equilibrium concentration of I?

21.Be sure to answer all parts.

The dissociation of molecular iodine into iodine atoms is represented as
I2(g) 2I(g)
At 1000 K, the equilibrium constant Kc for the reaction is 3.80 ×105. Suppose you start with 0.0461 mol of I2 in a 2.27L flask at 1000 K. What are the concentrations of the gases at equilibrium?

What is the equilibrium concentration of I2?

What is the equilibrium concentration of I?

Explanation / Answer

initial [I2] = number of mol / volume
= 0.0461 mol / 2.27 L
= 0.0203 M

I2 (g)   <---> 2I (g)
0.0203          0      (initial)
0.0203-x        2x      (at equilibrium)

Kc= [I]^2 / [I2]
3.80*10^-5 = (2x)^2 / (0.0203-x)
since Kc is small, x can be ignored as compared to 0.0203
Above expression thus becomes,
3.80*10^-5 = (2x)^2 / (0.0203)
(2x)^2 = 7.714*10^-7
x = 4.39*10^-4 M

[I2] = 0.0203-x = 0.0203 - 4.39*10^-4 = 0.0199 M
[I] = 2x= 2*4.39*10^-4 = 8.78*10^-4 M

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