The answers are provided below, I just want to know how they\'re found. 1. A qua

Question

The answers are provided below, I just want to know how they're found.

1. A quantity of HI was sealed in a tube, heated to 425°C, and held at this temperature until equilibrium was reached. The concentration of HI in the tube at equilibrium was found to be 0.0706 moles/liter. Calculate the equilibrium concentration of H2 (and I2). (Hint: H2 and I2 are equal).

For the reaction:                                 H2(g) + I2(g) ßà 2 HI(g)                  Kc = 54.6 at 425°C.

ans. 9.55 × 10-3M

2. Consider the reaction: 2 NO2(g) ßà N2O4(g).

A reaction is initiated at 100 °C with 2.35 moles NO2 in a 3.0 L reactor (no N2O4 present). If the equilibrium concentration of N2O4 is 0.25 M, What is the value of Kc at 100°C? (Hint: Make ICE table.)

ans. 3.2

Explanation / Answer

1. A quantity of HI was sealed in a tube, heated to 425°C, and held at this temperature until equilibrium was reached. The concentration of HI in the tube at equilibrium was found to be 0.0706 moles/liter. Calculate the equilibrium concentration of H2 (and I2). (Hint: H2 and I2 are equal).

For the reaction:                                 H2(g) + I2(g) ßà 2 HI(g)                  Kc = 54.6 at 425°C.

ans. 9.55 × 10-3M

Solution :-

H2 + I2 ------ > 2 HI                 Kc = 54.6

Kc equation is as follows

Kc= [HI]^2/[H2][I2]

We know the equilibrium concentration of the HI

So lets put it in the formula

54.6 = [0.0706]^2 / [x][x]

54.6 = [0.0706]^2 / x^2

Taking square root of both sides we get

7.389 = 0.0706 / x

x= 0.0706 M / 7.389

x= 0.00955 M

So the equilibrium concentration of the H2 and I 2 = x = 0.00955 M that is 9.55*10^-3 M

2. Consider the reaction: 2 NO2(g) ßà N2O4(g).

A reaction is initiated at 100 °C with 2.35 moles NO2 in a 3.0 L reactor (no N2O4 present). If the equilibrium concentration of N2O4 is 0.25 M, What is the value of Kc at 100°C? (Hint: Make ICE table.)

ans. 3.2

Solution :- initial concentration of the NO2 = 2.35 mol / 3.0 L = 0.7833 M

2 NO2(g) --------- > N2O4(g)

0.7833 M                         0

-2x                                      +x

0.7833 -2x                          0.25 M

Equilibrium concentration of the N2O4 = 0.25 M

So x= 0.25 M

Then equilibrium concetration of the NO2 = 0.7833 – 2x = 0.7833 – (2*0.25) = 0.2833 M

Equilibrium constant expression is as follows

Kc= [N2O4]/[NO2]^2

Kc =0.25 / [0.2833]^2

Kc = 3.53

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