[student8607]

When 25.0mL of 1.0M H₂SO₄ is added to 50.0mL of 1.0M NaOH at 25.0 degrees C in a calorimeter, the temperature of the aqueous solution increases to 33.9 degrees C. Assuming that the specific heat of the solution is 4.18K/g x degrees C, that its density is 1.00g/mL, and that the calorimeter itself absorbs a negligible amount of heat, calculate the  :delta: H (kJ) for the reaction.

H₂SO₄(aq) + 2NaOH(aq) -->2H₂O(l) + Na₂SO₄(aq)


I know the specific heat = heat involved / mass times change in temp
And that under a constant pressure, q = :delta: H,
but I'm not sure how to account for the mL and M?

[ARGOS++]

Dear student8607;

You need them to check if all of both will react, or if there is a limiting reagent, what would influence the whole amount of reaction. Check it with stoichiometry.

Good Luck!
                    ARGOS⁺⁺

[student8607]

Dear student8607;

You need them to check if all of both will react, or if there is a limiting reagent, what would influence the whole amount of reaction. Check it with stoichiometry.

Good Luck!
                    ARGOS⁺⁺

If my memory serves NaOH will be the limiting reagent.
So I just use that info?

[ARGOS++]

Dear student8607;

No!,  -  The Stoichiometry will tell you something different!.

Use your RxN:  1H₂SO₄ + 2NaOH -->2H₂O + Na₂SO₄
So how much NaOH is required?:  25ml * 1.0 molar / 1 =  ??  ml * 1.0 molar / 2

Calculate it and be happy!

Good Luck!
                    ARGOS⁺⁺

[student8607]

Ohh. I see. So the H2SO4 is the limiting reagent.

Dear student8607;

No!,  -  The Stoichiometry will tell you something different!.

Use your RxN:  1H₂SO₄ + 2NaOH -->2H₂O + Na₂SO₄
So how much NaOH is required?:  25ml * 1.0 molar / 1 =  ??  ml * 1.0 molar / 2

Calculate it and be happy!

Good Luck!
                    ARGOS⁺⁺

[ARGOS++]

Dear student8607;

Be happier!,  -  Neither is limiting!  -  All is just well balanced!

Good Luck!
                    ARGOS⁺⁺

[student8607]

Dear student8607;

Be happier!,  -  Neither is limiting!  -  All is just well balanced!

Good Luck!
                    ARGOS⁺⁺

Hmmm?
OK, so how would I set up this problem?

[ARGOS++]

Dear student8607;

Now you can do it exactly as you told before:    q = m * c * ∆T.
But remember that the result is only for the reaction in the given conditions/concentrations!

Good Luck!
                    ARGOS⁺⁺

[student8607]

OK, and do I just multiply liters by molarities to get moles and then convert to grams?

[ARGOS++]

Dear student8607;

No.  -  Density is given, so you can convert directly ml into grams!
Think about that the heat is finally stored in the water! For q molarity is of no interest.

Good Luck!
                    ARGOS⁺⁺

[student8607]

q = (4.18J/g x degrees C) (75g) (33.9 degrees C)
q = 10,627J

And the  :delta: H = q since they are at constant pressure, correct?

[ARGOS++]

Dear student8607;

Not exactly!,  -  You forgot: ∆T and not just T.
Otherwise it would be correct.

Good Luck!
                    ARGOS⁺⁺

[student8607]

q = (4.18J/g x degrees C) (75g) (8.9 degrees C)
q = 2,792.82J

And the  :delta: H = q since they are at constant pressure, correct?

[ARGOS++]

Dear student8607;

Ok!,  -  And think about exo- or endo-thermic (sign).

Good Luck!
                    ARGOS⁺⁺

[student8607]

it will be a negative sign because the surroundings gain heat