Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: KYR_Singularity on February 11, 2014, 03:13:53 PM
-
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
-
See: Hess's Law. Theres your hint….now lets see some work :)
Zack
-
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 :D
-
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.
-
Borek is correct. Otherwise, you got it!
-Zack
-
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?
-
(heats of formation of products) and subtract (heats of formation from reactants). Thank you Mr. Hess…..may I have another? :). Heats of formation.
-Z