Balanced Equations

[MedPR]

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Help please.

According to the following balanced reaction, the concentration of X with respect to Y is:

Reaction: nZ ---> (X)n + nY

A. n[Y]
B. n^2[Y]
C. (1/n)[Y]
D. (1/n^2)[Y]

Answer and explanation in white: C. In the equation the stoichiometric coefficient of X is 1 and the stoichiometric coefficient of Y is n. Therefore, the concentration of X is 1/n times the concentration of Y.

 

[phaloz]

Help please.

According to the following balanced reaction, the concentration of X with respect to Y is:

Reaction: nZ ---> (X)n + nY

A. n[Y]
B. n^2[Y]
C. (1/n)[Y]
D. (1/n^2)[Y]

Answer and explanation in white: C. In the equation the stoichiometric coefficient of X is 1 and the stoichiometric coefficient of Y is n. Therefore, the concentration of X is 1/n times the concentration of Y.

Ok it's basically an algebra problem.

For the equation aA + bB = cC + dD

The rate's for each specie is related using this general equation:
rate = (-1/a)[dA/dt] = (-1/b)[dB/dt] = (1/c)[dC/dt] = (1/d)[dD/dt]

'd' means change
't' is time
'a' 'b' 'c' and 'd' are the coefficients for each species
'A', 'B', 'C' and 'D' are the particular species

For our purposes we don't need dt and can just leave it has equal to 1.

So applying the formula to your equation we solve for X with respect to Y.

(1/1)X = (1/n)Y
Solve for x:
X= (1/n(1/1))Y = (1/n)Y

Hope that helps. If anything isn't clear let me know.

 

[typicalindian]

Ok it's basically an algebra problem.

For the equation aA + bB = cC + dD

The rate's for each specie is related using this general equation:
rate = (-1/a)[dA/dt] = (-1/b)[dB/dt] = (1/c)[dC/dt] = (1/d)[dD/dt]

'd' means change
't' is time
'a' 'b' 'c' and 'd' are the coefficients for each species
'A', 'B', 'C' and 'D' are the particular species

For our purposes we don't need dt and can just leave it has equal to 1.

So applying the formula to your equation we solve for X with respect to Y.

(1/1)X = (1/n)Y
Solve for x:
X= (1/n(1/1))Y = (1/n)Y

Hope that helps. If anything isn't clear let me know.

Holy crap lol..I just thought "for every X you get nY so for every Y you get 1/n X. Maybe my brain just severely over simplified it?

 

[phaloz]

Holy crap lol..I just thought "for every X you get nY so for every Y you get 1/n X. Maybe my brain just severely over simplified it?

For every x you would get

(1/1)X = (1/n)Y
(solve for Y)
(1/1)/(1/n)X = Y
(when you divide a fraction by a fraction you can treat it as a multiplication of two fractions. Just flip the denominator and numerator of the bottom fraction.)
nX = Y

 

[typicalindian]

For every x you would get

(1/1)X = (1/n)Y
(solve for Y)
(1/1)/(1/n)X = Y
(when you divide a fraction by a fraction you can treat it as a multiplication of two fractions. Just flip the denominator and numerator of the bottom fraction.)
nX = Y

oh haha thanks but I understood what was going on in the problem 😛 I was just saying that your explanation was really good for the OP. 👍

 

[gettheleadout]

I'd like to point out that this is confusing to read on SDN Mobile. I don't know if the n next to the X is superscript, but it just looks a coefficient to me, so I was like "uh...one X per Y?" lol

 

[pfaction]

I have no idea what's going on in this topic. What? I understand the equation aA + bB = cC + dD and then everything after that is like ???

 

[phaloz]

I have no idea what's going on in this topic. What? I understand the equation aA + bB = cC + dD and then everything after that is like ???

It's a reaction rate problem.

http://en.wikipedia.org/wiki/Reaction_rate#See_also


Check out the wikipedia link if you're confused. Near the top of the page it explains in detail about the formula I used. Once you understand the formula it basically becomes an algebra problem we're you are solving for a variable.

 

[pfaction]

It's a reaction rate problem.

http://en.wikipedia.org/wiki/Reaction_rate#See_also


Check out the wikipedia link if you're confused. Near the top of the page it explains in detail about the formula I used. Once you understand the formula it basically becomes an algebra problem we're you are solving for a variable.

Nope, I read it and I don't understand.

aA + bB = cC + dD
Reaction: nZ ---> (X)n + nY
Therefore, Xn = nZ-nY
so X should equal Z-Y??

 

[typicalindian]

It's a reaction rate problem.

http://en.wikipedia.org/wiki/Reaction_rate#See_also


Check out the wikipedia link if you're confused. Near the top of the page it explains in detail about the formula I used. Once you understand the formula it basically becomes an algebra problem we're you are solving for a variable.

reaction rate? I thought it was a stoichiometric problem? At least thats how I thought of it when I saw it

 

[phaloz]

Nope, I read it and I don't understand.

aA + bB = cC + dD
Reaction: nZ ---> (X)n + nY
Therefore, Xn = nZ-nY
so X should equal Z-Y??

First I want to point out that I assumed the (X)n was a typo of some sort. Looking at it doesn't make sense, especially when I read the answer to the problem. I would have rather written it like this for more clarity:
nZ ---> X + nY

zZ -----> xX + yY

rate = (-1/z)[dZ/dt] = (1/x)[dX/dt] = (1/y)[dY/dt]
Z,X,Y = concentrations of the species in Molarity
z,x,y = coeffecients in the reaction above

The parts of the equation that have [d(Some letter)/dt] is average rate of change for the particular species. The reason there is a negative sign in front of the reactants is because over time the reactions decrease and the rate of change for the reactants is negative, but the rate of reaction can't be negative so we get rid of negative sign.

Here are a couple of applications of the formula for 2 different reactions:
Here's the first:
N2 + 3 H2 → 2 NH3
rate = (-1/1)[d[N2]/dt] = (-1/3)[dH2/dt] = (1/2)[dNH3/dt]

Now for
4NH + 5O2 ----> 4NO + 6H2O
rate = (-1/4)[dNH/dt] = (-1/5)[dO2/dt] = (1/4)[dNO/dt] = (1/6)[dH2O/dt]

For this equation lets find the rate of reaction (rate) for NO in terms of H2O:
(1/4)[dNO/dt] = (1/6)[dH2O/dt]
divide both sides by (1/4)
[dNO/dt] = ((1/6) / (1/4)) [dH2O/dt]
[dNO/dt] = (2/3) [dH2O/dt]
Since the equations are equal to each other all I had to do was solve for [dNO/dt].

Does that help clear things up?

 

[phaloz]

reaction rate? I thought it was a stoichiometric problem? At least thats how I thought of it when I saw it

Well in a sense you're right because you do use the stoichiometric coefficients when solving.

 

[milski]

X👎 is a molecule of n X atoms, as H2.

Not sure why you have to involve rates at all, since it's purely stoichiometric problem you get n moles of Y for each mole of X.

You cannot really talk about rates of the reaction since you don't know the mechanism. But whatever they are, the stoichiometry at the end does not change.

 

[phaloz]

X👎 is a molecule of n X atoms, as H2.

Not sure why you have to involve rates at all, since it's purely stoichiometric problem you get n moles of Y for each mole of X.

You cannot really talk about rates of the reaction since you don't know the mechanism. But whatever they are, the stoichiometry at the end does not change.

I was thinking same as you. Just set X = nY, however if you look at the answer provided in the OP its (1/n).

 

[MedPR]

X👎 is a molecule of n X atoms, as H2.

Not sure why you have to involve rates at all, since it's purely stoichiometric problem you get n moles of Y for each mole of X.

You cannot really talk about rates of the reaction since you don't know the mechanism. But whatever they are, the stoichiometry at the end does not change.

That's what I thought too, but the answer is C.

The concentration of X with respect to Y is 1/n[Y]...

 

[gettheleadout]

So what exactly is "(X)n" trying to represent in the OP? Or is it a typo?

 

[milski]

I was thinking same as you. Just set X = nY, however if you look at the answer provided in the OP its (1/n).

X=nY would be the opposite - it's n moles of X for 1 mole of Y. The correct equation is nX=Y or X=1/n * Y

That's what I thought too, but the answer is C.

The concentration of X with respect to Y is 1/n[Y]...

Exactly - for each mole of X you get n moles of Y. Y/X=n or Y=nX or X=1/n Y

So what exactly is "(X)n" trying to represent in the OP? Or is it a typo?

I'm fairly sure it's the number of X atoms in a molecule of the substance X. From example:
N3O3 -> O3 + N3 - here 3 is n, X is O (oxgen) and Y is N (nitrogen). The reaction is made up, of course.

 

[chiddler]

Why was it written as "(X)n" while Z and Y were written "nY" and "nZ"?

 

[gettheleadout]

Why was it written as "(X)n" while Z and Y were written "nY" and "nZ"?

By what milski is saying, it's because the n isn't a coefficient in this case but a subscript.

 

[chiddler]

"Y=nX"

so if i'm understanding correctly, question indicates that the coefficient of X is 1. Coefficient of Y is n.

x = Yn

how is the Y=nX set up?

What significance does n have with X? Why is the subscript important?

 

[milski]

The n subscript is not that significant although the equation would look weird without it.

Think about a decomposition of something:

A -> X + nY

That means that from one mole of A you get one mole of X and n moles of Y. Thus [Y]=n[X]

 

[MedPR]

The n subscript is not that significant although the equation would look weird without it.

Think about a decomposition of something:

A -> X + nY

That means that from one mole of A you get one mole of X and n moles of Y. Thus [Y]=n[X]

This did it for me. To everyone wondering, the OP is not a typo. And yes, the n subscript on the X term does make it confusing (I think that was the point of the question).

So in A--> X + nY, say n=5.

That means for every 1mol of X formed, you also get 5mols of Y.

So algebraically 5X=1Y (this algebra stuff has really gotten me twisted around on quite a few problems)

So from 5X=1Y, you get X=Y/5, or X=(1/5)Y

 

[MedPR]

"Y=nX"

so if i'm understanding correctly, question indicates that the coefficient of X is 1. Coefficient of Y is n.

x = Yn

how is the Y=nX set up?

What significance does n have with X? Why is the subscript important?

The n subscript on the X is like (NH3)3. Three NH3s, versus 3NH3 = 3moles of NH3.