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11 Nov 2019

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
×
10−5. Suppose you start with 0.0451 mol of I2 in a 2.25−L flask at 1000 K. What are the concentrations of the gases at equilibrium?

What is the equilibrium concentration of I2?

M

What is the equilibrium concentration of I?

M

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Trinidad Tremblay
Trinidad TremblayLv2
10 Jan 2019

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Related questions

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 × 10−5. Suppose you start with 0.0454 mol of I2 in a 2.28−L flask at 1000 K. What are the concentrations of the gases at equilibrium?

What is the equilibrium concentration of I2? M

What is the equilibrium concentration of I? M

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

×

10−5. Suppose you start with 0.0461 mol of I2 in a 2.27−L flask at 1000 K. What are the concentrations of the gases at equilibrium?

What is the equilibrium concentration of I2?

M

What is the equilibrium concentration of I?

M

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 × 10−5. Suppose you start with 0.0460 mol of I2 in a 2.32−L flask at 1000 K. What are the concentrations of the gases at equilibrium? I2 and I

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