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So I'm doing some chemistry review for my exam in a couple weeks and I'm having some problems getting the right answer because of the signs that should be used.

Eg. HNO3 + KOH > KNO3 + H2O    DeltaH = -53.4 kJ/mol
55.00mL of 1.3M solutions of both reactants @ 21.4 degree C is mixed in a calorimeter. What is the final temp of the mixture, assume density of both solutions is 1.00g/mL adnd c = 4.184 (that of for water)

Q= (-53.4)
m=110g
c=4.184
t= (t2-21.4)

My Solution

Q=mct
(-53.4) = 110 x 4.184 x (t2 - 21.4)
(-53.4) = 460.24t2 - 9849.136
(-53.4) + 9849.136 = 460.24t2
t2 = 21.28

But the answer is wrong, it should be around 30 degrees.

I dont know what I'm doing wrong.

Im sort of questioning whether I should get away from Q=mct and use Qreactant= -Qsolution,
but I dont understand how to use that exactly.

Couild anyone help me with the formula that I'm using, or what I'm doing wrong, because most of my questions are wrong because of this.

It was so easy four months ago...

[enahs]

First, units units units.
Your ?H is in kJ but your specific heat is in joules.

Also, your ?H is for the equation as written, which happens to be 1:1 mole ratio for reactants. While you do have a 1:1 mole ratio for reactants, you do not actually have 1 mole of either one of them.

So once you find out how many moles you have, you multiply that times the -53.4kJ to find out how many kJ are released in the reaction with that amount of moles.

Then you solve just like you have set up (with the correct units, both in the same unit). But you do not use -X on the left hand side (where you have -53.4), because as you said (though slightly wrong, reaction, no reactants) Q_rxn=-Q_solution. You are asked for the temperature of the final solution.

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I did the same process like you told me but I still didnt right the right answer, here's what I did:

Find Mols HNO3

55g x 1 mol/63g x 1L / 1.3mol x 1 mol/22.4L = 0.03 mols

Find Mols KOH

55 x 1mol/56g x 1L/1.3mol x  1mol/22.4L = 0.034 mol

Add them together you get 0.064 mols x -53.4 = -3.42 kJ...then

Qrxn = -Qsoln
110 x 4.184 x (t2 - t1) = -(-3.42)
460.24t2 - 9849.136 = 3.42
t2 = 21.4

[enahs]

No, you missed the very first thing I said. UNITS!!!!!. Along with what you did, you must convert the units!
The enthalpy is in kJ while the specific heat is in joules. That will get you closer.

You are not also converting to moles properly/are introducing unneeded error.

Molarity = moles per liter.
If you multiply the molarity (M) times the volume in liter, you get the number of moles.
So that becomes 55/1000 * 1.3 or
0.055 L * 1.3 mol/L = 0.0715 moles.

[Yggdrasil]

Find Mols HNO3

55g x 1 mol/63g x 1L / 1.3mol x 1 mol/22.4L = 0.03 mols

Find Mols KOH

55 x 1mol/56g x 1L/1.3mol x  1mol/22.4L = 0.034 mol

Add them together you get 0.064 mols x -53.4 = -3.42 kJ...then

In additon to what enahs said about units, this part is wrong also.  The reaction produces 53.4kJ of heat when 1 mol of HNO3 with 1 mol of KOH.  So when 0.03 mol of HNO3 reacts with 0.034 mol  of KOH, you do not get 0.064 * 53.4 kJ of heat.  You only get 0.03 * 53.4 kJ of heat (since HNO3 will be the limiting reagent).

[///]

Ughh Im still not getting it right, even after all those mistakes fixed.

Could you maybe explain to me the process of using Q=mct because I think thats where I'm getting it wrong, just figuring out when to use plain old Q=mct and when to use mct = -mct;;

And isnt the limiting reaction the one with the least mass, right? So then why was HNO3 the limiting reagent, when you figure out the # grams, it is 4.5g vs KOH with 4.0g.

[enahs]

No, a limiting reaction is one in which the moles of available reactants is less then the other (not mass), when compared to the stoichiometry of the balanced equation. In this case, you have the same volume and same concentration of both reactants, so they both have the same number of moles (1:1). And because the stoichiometry is 1:1, this is not a limiting reagent problem.

Molarity = Mols
                     L


Mols   *    L    =   Mols
      L

L = an L that is crossed out.


This Q = -Q, or as you wrote the mathematical equivalent mct = -mct, is missing something.

That is Q_Reaction = -Q_Solution
You are not identifying that they are for different conditions.

This comes from a simplification for the math at your level. This assumption is that there is no heat lost to the environment, container, etc etc. This is impossible in real life of course.
But what this means is that Q1 + Q2 …= 0

In this case:
Q_rxn + Q_solution = 0
Or when you bring one to the other side
Q_rxn = -Q_solution

Your question specifically ask you to calculate the enthalpy of the solution. But the value you are given is the enthalpy of the reaction. Making the assumptions at your level that you are allowed to make, then the enthalpy of the solution is the same as the enthalpy of the reaction, just opposite sign.

And units unit units. C, the specific heat is in joules (typically, unless otherwise stated). Enthalpy is in kilojoules. They must be the same in order to use Q= C • m • ?T

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So how would you solve the equation ?

The answer is 29.7 degrees, however still with your explanation (thank-you for that), I get the same answer of 21.3 degrees

Could you possibly solve it for me showing me all the steps involved?

Thanks in Advance

[enahs]

(55mL) * ( 1 L       )   *   ( 1.3 mol )    = 0.0715 mol (of both HNO₃ and KOH)
               1000 mL                  L

Because there is the same number of moles (by definition of being the same molarity and same amount of solution) they are in a 1:1 ratio, just like you balanced equation calls for.

(0.0715 mol) * (-53.4 kJ) = 3.8181 kJ
                           mol

Does this number make sense, yes, think about it.
0.0715 moles is 7.15% of 1 mole. 7.15% of -53.4 = 3.8181.
This is the enthalpy for that many moles of the reaction.

You are asking to calculate the Q for the solution, and because of the assumptions we talked about earlier, Q_rxn = -Q_solution : this negative can go on either side and it changes nothing about it mathematically. It just means same value but opposite sign.
So Q_solution = -(-3.8181) which = +3.8181 kJ

Then with your equation (do you get why this equation works? cancel the units on the right and you are left with just energy, same on the left):
Q = m * c * (T₂-T₁)

3.8181 kJ = 110 g * 0.004184   (   kJ     ) * (x - 21.4)
                                                    (  g * K  )
or
3818.1 J = 110 g * 4.184   (    J    )* (x - 21.4)
                                            ( g * K )

x = 29.69588910 = 29.7 with sig figs.

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Thanks a LOT; this clarifies a lot for me;;

Just a couple questions

We got

(0.0715 mol) * (-53.4 kJ) = 3.8181 kJ
                           mol

So we got 3.8181kJ = 3818J

but then over here

So Q_solution = -(-3.8181) which = +3.8181 kJ

Why did it become negative.
If Qrxn = -Qsoln and 3818 = Qrxn then 3818 = -Qsoln ;;; or -3818 = Qsoln,

[enahs]

You are welcome

As I said in the last post.

"this negative can go on either side and it changes nothing about it mathematically. It just means same value but opposite sign."

In other words.

x=-x
But if x as actually (-x)
Then you have (-x)=-(-x) or -x=x (this is the exact same thing as x=-x)

This Q = -Q is just a inverse relationship. When one positive the other is negative.

And think about your assumptions, that no heat is lost to the environment or comes from the environment (the assumption is that it is perfectly insulated). If one loses heat the other must gain it. If one gains heat the other must lose it. Those are equal in magnitude but opposite in direction.

You are mathematically thinking of the - sign to be there; it goes in front of whatever number you replace Q with. You can replace Q with a negative number and you then have two negatives (or a positive).

In this case, Qrxn = -3.8181 kJ
Qrxn = -Qsol, multiply both side by -1
-Qrxn = Qsol
-Qrxn = -(-3.8181 kJ) = +3.8181 kJ