hw question #3

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Kaylee Nezwek 1D
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hw question #3

Postby Kaylee Nezwek 1D » Sat Jan 16, 2021 3:13 pm

Hi! For number 3 on the homework, it says, " At a certain temperature, H2 + I2 -> 2HI, Kc = 53.3, at this temperature, 0.700 mol H2 and 0.700 mol I2 were placed in a 1.00 L container to react. What concentration of HI is present at equilibrium?" I set up Kc to be:
53.3 = (2x)^2/(0.7 - x)(0.7 - x)
But I'm a bit confused on how to get the quadratic equation from this. Can someone explain how to get the quadratic equation/what it would be?

Brennan McGurrr 3C
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Re: hw question #3

Postby Brennan McGurrr 3C » Sat Jan 16, 2021 3:19 pm

You would have to FOIL the denominator then multiply that expression to the other side. This will allow you to set the equation equal to 0 and use the quadratic formula. If you have a graphing calculator, you can graph both sides of the equation and find where they intersect to solve this. I find this method to be faster and easier. Desmos is a good website for an online graphing calculator.

Nathan Lao 2I
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Re: hw question #3

Postby Nathan Lao 2I » Sat Jan 16, 2021 3:20 pm

For this specific K equation, you can actually take the square root of both sides. It would simplify to square root of 53.3 = 2x/(0.7-x) and then you would solve for x from there by rearranging the equation.

LexyDenaburg_3A
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Re: hw question #3

Postby LexyDenaburg_3A » Sat Jan 16, 2021 4:56 pm

In this question, how did you set up Kc?

Sophia Wendin 3L
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Re: hw question #3

Postby Sophia Wendin 3L » Sat Jan 16, 2021 5:36 pm

multiply out the denominator( (a + b)^2 = a^2 + 2ab + b^2) then cross multiply with 53.3. Move everything to one side so you can find X using the quadratic formula.

Lung Sheng Liang 3J
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Re: hw question #3

Postby Lung Sheng Liang 3J » Sat Jan 16, 2021 6:05 pm

This calculation is simple enough to the point where you don't need to utilize the quadratic formula. You can get the square root of both sides and then solve for x and then to get the concentration for HI multiply that value by 2.

Shruti Kulkarni 2I
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Re: hw question #3

Postby Shruti Kulkarni 2I » Sat Jan 16, 2021 6:10 pm

You can multiply both sides by (0.7 -x)^2 to remove the denominator and then expand each expression to create an equality. Afterward, you can move all the terms to one side and set the equation equal to zero. All that's left is to solve using the quadratic equation or even graphing out the equation to find the zeros.

Alisa Nagashima 1B
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Re: hw question #3

Postby Alisa Nagashima 1B » Sun Jan 17, 2021 1:00 am

Hi! Essentially, you just need to rearrange the equation so that it fits the ax2 + bx + c = 0 form since it makes it easier to use the quadratic formula. But since the components of the denominator are the same for this equation, it's easier to take the square root of both sides.

Kaleb Tuliau 3E
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Re: hw question #3

Postby Kaleb Tuliau 3E » Sun Jan 17, 2021 2:07 am

for question 3 when does Oxygen's given equilibrium mole play a role? (if this question makes sense)

Uyen Trinh 3C
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Re: hw question #3

Postby Uyen Trinh 3C » Sun Jan 17, 2021 2:19 am

What helps me get at these chemical equilibrium problems is handwriting everything out. Here is how I did it as well as the way Sapling teaches us to do it. I erased the answers but left all my workings. I hope this helps.

Image

Gian Boco 2G
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Re: hw question #3

Postby Gian Boco 2G » Sun Jan 17, 2021 3:32 am

On the right side of the equation, both the numerator and the denominator have a square (once you combine the like terms in the denominator), meaning the entire fraction can be squared. That way, you can square root both sides of the equation and solve with algebra from there.


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