Manipulating chemistry equation to form quadratic equation

[clarkin]

Dear Colleagues,

I have been introduced to the equilibrium constant expression in Chemistry and have been attempting a question on manipulating it to form a quadratic equation, so that it can be solved.

The chemistry equation I'm attempting to manipulate is:


X^2
1.80M = ------------- and to convert this to the form ax^2 + bx + c = 0
0.300 - X

I have been attempting to do this by trying to isolate x^2 by itself by multiplying both sides by 0.300 -x but still get the wrong answer. Any solution is appreciated.

Is there anything on google that can instruct on how to do this ?.

Thank you.

 

[Deleted member 4993]

Dear Colleagues,

I have been introduced to the equilibrium constant expression in Chemistry and have been attempting a question on manipulating it to form a quadratic equation, so that it can be solved.

The chemistry equation I'm attempting to manipulate is:


X^2
1.80M = ------------- and to convert this to the form ax^2 + bx + c = 0
0.300 - X

I have been attempting to do this by trying to isolate x^2 by itself by multiplying both sides by 0.300 -x but still get the wrong answer. Any solution is appreciated.

Is there anything on google that can instruct on how to do this ?.

Thank you.

Does your equation look like:

X^2 + 1.80 * M = 0.300 - X

Please also tell us

What is 'M'?

Is it a constant relative to 'X'?

 

[clarkin]

Does your equation look like:

X^2 + 1.80 * M = 0.300 - X

Please also tell us

What is 'M'?

Is it a constant relative to 'X'?

Dear Subhotosh, thank you for responding. 'M' is the concentration

The equation is: 1.80M = X^2 and to manipulate this so that it resembles ax^2+bx+c=0
------------
0.300 - X

The original question is:

We ask what the concentrations at equilibrium will be. The answer? They will be simply the sum in each column. That is, the amount of PCl₅ will be the amount we started with, minus the amount that reacted. The amounts of the proudcts will be the amounts we started with plus the amounts produced. Add a last row to the table:

	PCl5	PCl3	Cl2
Start	[ 0.300 ]	0	0
Change	- x	x	x
At Equilibrium	[ 0.300 - x ]	[ 0 + x ]	[ 0 + x ]

 

[clarkin]

Denis, thank you for responding.

the equilibrium expression to manipulate so that it resembles ax^2 + bx + c =0 is:

1.80 = x^2
_________
0.300 - x

The answer is supposed to be: x^2 +1.80x - 0.540 = 0

I've tried the normal algebraic manipulation by attempting to get x^2 by itself by multiplying both sides
by '0.300 - x' but get the wrong answer. Also, why does the 1.80 number end up with 'x' next to it ?.

 

[Deleted member 4993]

Denis, thank you for responding.

the equilibrium expression to manipulate so that it resembles ax^2 + bx + c =0 is:

1.80 = x^2
_________
0.300 - x

The answer is supposed to be: x^2 +1.80x - 0.540 = 0

I've tried the normal algebraic manipulation by attempting to get x^2 by itself by multiplying both sides
by '0.300 - x' but get the wrong answer. Also, why does the 1.80 number end up with 'x' next to it ?.

Are you sure you wrote that equation correctly? Please check it over and confirm.

 

[clarkin]

Subhotosh, apologies - this is the question with answer:

Step 6 Now plug the concentrations from the last row into the equilibrium expression.

A little algebraic manipulation leads to a quadratic equation:

What I don't understand is, how the figure '1.80' has an x next to it in the answer and that the equation is equal to zero ?.

 

[stapel]

Subhotosh, apologies - this is the question with answer:

Step 6 Now plug the concentrations from the last row into the equilibrium expression.

A little algebraic manipulation leads to a quadratic equation:

What I don't understand is, how the figure '1.80' has an x next to it in the answer and that the equation is equal to zero ?

To learn how to solve rational equations, try here. Once you've studied the basic terms and techniques, please return to the exercise:

. . . . .\(\displaystyle 1.80\, =\, \dfrac{x^2}{0.300\, -\, x}\)

You'll then understand that the first step in solving is to multiply through by the denominator, clearing the fractions. It will also explain why the "0.300" "became" "0.540".

 

[clarkin]

Dear Stapel,

Thank you for the link.
I have attempted the following question:


0.30 = x^2 / 0.2 -x

I cross multiplied and got this equation : x^2 = 0.30 ( 0.2 -x ), then multiplied everything inside the bracket:
x^2 = 0.06 - 0.30x

The answer is supposed to be : x^2 + 0.03x - 0.006 = 0 ( of the form ax^2 + bx + c = 0 )

Can someone please assist as to how we arrive at this answer ?.

Thank you

 

[mmm4444bot]

0.30 = x^2 / (0.2 - x)

I … got … x^2 + 0.30x - 0.06 = 0

The answer is supposed to be : x^2 + 0.03x - 0.006 = 0 Says who? :cool:

There might be a typo, somewhere in your materials.

 

[clarkin]

Dear mmm4444bot,

just checked the answer - there's no typo:

The only way I can think of getting that answer ( x^2 + 0.03x - 0.006 = 0 ) is to divide the 0.30x and 0.06 by 10 - but why ?

 

[Deleted member 4993]

Dear mmm4444bot,

just checked the answer - there's no typo:

The only way I can think of getting that answer ( x^2 + 0.03x - 0.006 = 0 ) is to divide the 0.30x and 0.06 by 10 - but why ?

Not the answer - but the typo might be in the original problem statement.