(a) In an experiment, 2.0 mmol Cl2 was sealed into a reaction vessel of volume 2.0 L and heated to 1000. K to study its dissociation into Cl atoms. Use the information in Table 5G.2 to calculate the equilibrium composition of the mixture.
Hello, I understand this question conceptually and I believe I'm setting up my equation correctly, but I keep getting the wrong answer for X according to the solution in the textbook (the textbook says X = 5.5x10^-6)
I set my equation up to find X like this: 1.2x10^-7 = 2x^2/0.001-x
I solve through the quadratic equation but I can't seem to get the textbook solution for X. I'm not sure what I'm doing wrong
Textbook 5i.13 part a
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Re: Textbook 5i.13 part a
Hi! I think you have the right setup as well. Nice job!
1.2x10^-7=2x^2/0.001-x
1.2x10^-7=4x^2/0.001-x
0=4x^2 + (12.x10^-7)x - (1.2x10^-10)
Then solve using the quadratic formula. -b±√b^2-4ac/(2a)
Hopefully these steps help a little!
1.2x10^-7=2x^2/0.001-x
1.2x10^-7=4x^2/0.001-x
0=4x^2 + (12.x10^-7)x - (1.2x10^-10)
Then solve using the quadratic formula. -b±√b^2-4ac/(2a)
Hopefully these steps help a little!
Re: Textbook 5i.13 part a
Hi, can someone explain why you would have 2x^2 in the numerator for the products? I thought it would just be X^2 from the molar ratio of the chemical equation.
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Re: Textbook 5i.13 part a
to answer the above question, it is because you must bring down the coefficient in the equation to the coefficient on the x
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Re: Textbook 5i.13 part a
Make sure that when you are looking at the table, you use the KC value and not the K value.
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Re: Textbook 5i.13 part a
Yes, there is 2Cl(g) in the equation meaning that for x decreasing for Cl2, there is an increase in 2x in Cl (molar ratio being 1:2). This applies to any other case where there are molecules in an equation that are multiplied by more than 1.
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