Asked by TOM

The heat evolved in the reaction of hydrobromic acid with potassium hydroxide was determined using a coffee coup calorimeter. The equation for the reaction is:

HBr(aq) + KOH(aq) -------------> KBr(aq) + H2O(l)

When 25.0 mL of 1.00 M HBr at a temperature of 24.5oC was quickly mixed with 25.0 mL of 1.00 M KOH, also at 24.5 oC, the temperature rose to 31.4 oC. Calculate the enthalpy of reaction for this chemical change. Assume that the specific heats of each solution a re 1.00 cal/g oC and that the densities of each solution are 1.00 g/mL.

The answer should be -13.8kcal/mol but I don't know how to get that answer.
Answered by DrBob
You have 25.0 mL x 1.00 M = 25 mmols or 0.025 mols HBr and KOH reacting.
How much heat is generated? q = mcdT. m = mass = 25 + 25 = 50 grams. c = 1 cal/g. dT = delta T = 31.4-24.5 = 6.9 degrees.so q = 345 cal per 0.025 mols. How much is that for 1 mol? Remember this is an exothermic reaction so dH will be negative. Post your work/thoughts if you have more trouble.
Answered by TOM
thank you very much
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