Can someone please help me on the following problem:
At a certain temperature, the equilibrium constant, Kc, for this reaction is 53.3.
H2(g)+I2(g) <--------> 2HI(g) Kc=53.3
At this temperature, 0.300 mol H2 and 0.300 mol I2 were placed in a 1.00 L container to react. What concentration of HI is present at equilibrium?
When I set up the ICE table and then I solved for x. In the end, I found that x= 0.413 and 0.235. BUt both of these numbers are positive so how do you determine which one is the correct x value?
Solving for the correct X value
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Re: Solving for the correct X value
You don't actually need to use the quadratic equation for this problem!
After setting up your ICE box, you should get Kc = (2x)2/(0.3-x)2. Since both the numerator and the denominator are squared, we can take the square root of both sides to get √Kc = (2x)/(0.3-x). Then, plug in your value for Kc and solve for x. I got 0.471 for x.
After setting up your ICE box, you should get Kc = (2x)2/(0.3-x)2. Since both the numerator and the denominator are squared, we can take the square root of both sides to get √Kc = (2x)/(0.3-x). Then, plug in your value for Kc and solve for x. I got 0.471 for x.
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Re: Solving for the correct X value
In general, if solving the quadratic results in two positive values for x, the correct value can be found by plugging x back into the expressions for the equilibrium concentrations. Even if x itself is not negative, if the initial concentration of one of the reactants minus x results in a negative value for the equilibrium concentration, that solution would be incorrect.
Hopefully this helps and please correct me if there are any mistakes.
Hopefully this helps and please correct me if there are any mistakes.
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Re: Solving for the correct X value
In this case, the initial concentration of H2 and I2 is 0.3. So x=0.413 couldn't be the x value, because in this case, the equilibrium concentration of H2 would be negative.
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