In: Chemistry
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
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