In: Chemistry
At a certain temp, the equilibrium constant for this reaction is 53.3. H2(g)+I2(g)=2HI(g) At this temperature, 0.7 mol of H2 and 0.7 mol of I2 were placed in a 1 L container. What is the concentration of HI present at equilibrium?
Since volume is 1 L value of concentration and number of moles will be same
H2 (g) + I2 (g)
<-------------------> 2HI (g)
0.7
0.7
0 (Initial)
0.7-x
0.7-x
2x (at equilibrium)
Kc = [HI]^2 / {[H2] [I2]}
53.3 = (2x)^2 / (0.7-x)^2
53.3 = (2x / (0.7-x))^2
2x / (0.7-x) = sqrt (53.3)
2x / (0.7-x) = 7.3
2x = 5.11 - 7.3*x
9.3*x = 5.11
x = 0.55 M
[HI] = 2*x
= 2*0.55 M
= 1.1 M
Answer: 1.1 M