[KYR_Singularity]

25.0 cm3 of 2.00 mol dm-3 HCl(aq) was mixed with 25.0 cm3 of 2.00 mol dm-3 NaOH(aq). The temperature increased from 22.5°C to 34.5°C. Find the enthalpy change of reaction for the following equation:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l).
 
how do you work out the energy change in this reaction ?
please show all the work out  , thank you

[zsinger]

See:  Hess's Law.  Theres your hint….now lets see some work
        Zack

[KYR_Singularity]

HCL + NaOH -> NaCl + H2O
(25+25) x 4.18 x 12 = 2508 J
25 Volume reacted
25 x2 / 1000 = 0.05 mol
2508/0.05=50.16KJ/mol
^H = -50.16KJ/mol
is this right ? thanks for the tip anyway 

[Borek]

Doesn't look bad, although you may want to take water produced to the total mass.

Some will consider it nitpicking, but it is almost a 2% difference.

[zsinger]

Borek is correct.  Otherwise, you got it!
           -Zack

[Benzene]

In Hess's Law, it states that the sum δH products-δH reactants = δH reaction
Is the δH of the products and reactants calculated from the δH of formation? or combustion?

[zsinger]

(heats of formation of products) and subtract (heats of formation from reactants).  Thank you Mr. Hess…..may I have another?   .  Heats of formation.
        -Z