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.

ΔH = -57.78kj/mol

Explanation:

Assumptions

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

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when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.