5I.19 A reaction mixture that consisted of 0.400 mol H2 and 1.60 mol I was introduced into a flask of volume 3.00 L and heated. At equilibrium, 60.0% of the hydrogen gas had reacted. What is the equilibrium constant K for the reaction H2(g) + I(g) ~ 2 HI(g) at this temperature?
I keep on getting K=1.03 but the answer key says it’s K=1.1, it’s a big difference so I’m assuming there’s something wrong in the way I approached this problem, does anyone know how to correctly do this ?
Textbook problem 5I.19
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Re: Textbook problem 5I.19
To solve this problem, I created an ICE chart with 0.133M and 0.533M as the initial concentrations of H2 and I2. I got this by dividing the moles of each reactant by 3L. The change in concentration can be determined by calculating the amount of hydrogen gas reacted. This can be done by multiplying the initial concentration of H2 by 0.6 since 60% of H2 was reactant. The change in concentration is roughly 0.008M. The rest of the ICE chart is then completed by subtracting the reactants by the change in concentration and adding 2 times the change in concentration to the product because for every 1 mole of H2 reacted, 2 moles of HI are produced. The equilibrium concentrations can then be determined and K is calculated by taking the equilibrium concentration of the product squared, and dividing that by the concentration of the reactants. I've also attached a picture of my work for extra clarity. Hope this helps!
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