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.
5 answers
This is Joe again the answer should be -13.8kcal/mol but i don't know how to this get this answer.
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 this get this answer.
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 this get this answer.
I answered this above. The answer of -14.8 kcal/mol is correct.
See post above with Tom.
That's a typo. It should be -13.8 kcal/mol. See above under post by Tom for my response.