H2(g) + I2(g) = 2HI(g)

asked by @jeremyb10 • 10 months ago • Chemistry • 5 pts

I am told to calculate the Molarity for I2.. I am told the volume of the flask is 250mL and there are 0.90molH2 and 0.70mol HI. I am given the equillibrium constant K which is equal to 8.84.

I know this is a problem that I am likely going to have to use an ICE chart for, as well as the quadratic formula. However, I get very confused once I get my values from the ICE chart and where to go from there.

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Hey there! You would have to be specific if mol values are before equilibrium or are the values when it is already in equilibrium? Since it is not specified I'll just assume that it is before equilibrium with [I2] is initially 0. enter image description here I hope this helps!.

answered by @leot1 • 10 months ago
  • Why is the x value for H2 and I2 negative? I thought we lose reactants to make products, and in this case H2 and I2 are reactants in the given reaction? Jeremy commented 10 months ago
  • Previous comment should have said WHY ISNT**** Jeremy commented 10 months ago
  • The answer you provided is not right, by the way. Jeremy commented 10 months ago