what is the equilibrium expression for the equation h2(g) + i2(g) == 2hi(g) Solu

Question

Explanation / Answer

First thing you have to do is take the moles and the liters and get the molarity for both the H2 and the I2. M=moles / liters M of H2= 2.00/1.06= 1.88 M H2 M of I2= 3.60/1.06= 3.39 M I2 Now you have the starting molarities for everything in the equation, 1.88M H2, 3.39M I2, and 0M HI. Next you do the ICE table, ICE stands for Initial, Change and Equilibrium. H2(g) + I2(g) ==> 2HI Initial 1.88 3.39 0 Change -x -x + 2x Equilibrium 1.88 - x 3.39 - x 2x K= [HI]^2/[H2][I2]=(2x)^2 /(1.88 - x)(3.39 - x) Now we assume that x is very small, close to zero, that it is not going to affect the number infront of it. Therefore: K= 4x^2/(1.88)(3.39) For some reason I did my calculations and they don't seem reasonable. I think that your equilibrium constant K was not written down correctly, but if you do end up getting the correct K then my steps should be fine. Just solve for x and then use it to subtract the x from the original concentration of I2 and then you will have the equilibrium concentration for I2. Let X = moles of H2 and I2 that disappear to form HI during the reaction. The equilibrium concentrations of H2 and I2 at equilibrium will be (2-X)/1.06 and (3.6-X)/1.06, and the concentration of HI at equilibrium will be 2X/1.06. That make sense so far? Now, just plug all that into your expression for K and solve for X. You'll see that the 1.06's will all just cancel out, and you'll have: 49.7 = (2X)^2/(2-X)(3.6-X) It won't be fun to solve, and you'll have to use the quadratic equation, but you should be able to do it...

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.