In an experiment to determine ΔH for the reaction between HBr and NaOH , 43.6 mL of 1.08 M HBr at 19.95 °C is placed in a coffee cup calorimeter which has a heat capacity of 7.99 J/°C. 43.6 mL of 1.08 M NaOH at 19.95 °C is added to the acid solution in the calorimeter and quickly mixed. A final temperature of 26.88 °C was recorded. What is ΔH for this reaction, in kJ? Assume that the specific heats of all solutions are the same as for pure water (4.18 J g-1 °C-1), and that all solution densities are 1.00 g/mL. Don't forget to put the proper sign on your ΔH value.

These are solutions, not pure water. The specific heat of water is 4.184 J/goC. Assume that these solutions are close enough to being like water that their specific heats are also 4.1984 J/goC.

the heat evolved will be

q = mcΔt

first of all convert ml to grams

(43.6 mL + 43.6 mL ) = 87.2mL of solution.

density of water=1g/ml

87.2 mL X 1 g/ml = 87.2 grams of solution.

(mass = Volume X Density)

Find the temperature change.

Δt =tfinal - tinitial = 26.88oC - 19.95oC = 6.93oC

q = mcΔt

= 87.2 grams X 4.184 J/goC X 6.9oC

= 2.52 X 10^3 J

This is the heat gained by the water, but in fact it is the heat lost by the reacting HBr and NaOH, therefore q = -2.52 x 10^3 J.

i.e. it is an exothermic reaction, heat was lost to the water and it got warmer

to find the how much of HBr that was used in mol

43.6 mL of HBr X 1.00 mol HBr/ 1000 mL HBr = 0.0436 mol HBr

same quantity of base NaoH was used

. molar enthalpy = J/mol = -2.52 x 10^3 J / 0.0436 mol

-57.78kj/mol

Therefore, for the neutralization of HBrl and NaOH, the enthalpy change, often called the enthalpy of reaction is ΔH = -57.78kj/mol