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In a coffee-cup calorimeter, 1 mol NaOH and 1 mol HBr initially at 28 oC (Celsius) are mixed in 100g of water to yield the following reaction: NaOH + HBr → Na+(aq) + Br-(aq) + H2O(l) After mixing the temperature rises to 88.5 oC. Calculate the change in enthalpy of this reaction. Specific heat of the solution = 4.184 J/(g oC) State your answer in kJ with 3 significant figures. Don't forget to enter the unit behind the numerical answer. The molecular weight of NaOH is 40.0 g/mol, and the molecular weight of HBr is 80.9 g/mol. ΔH =

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Answer:

The enthalpy change during the reaction is -56.0 kJ/mole.

Explanation:

Mass of 1 mole of NaOH = 40.0 g

Mass of 1 mole of HBr = 80.9 g

Mass of water = 100 g

Mass of solution ,m = 100 g + 40.0 g + 80.9 g = 220.9 g

First we have to calculate the heat gained by the solution.

[tex]q=mc\times (T_{final}-T_{initial})[/tex]

where,

m = mass of solution

q = heat gained = ?

c = specific heat = [tex]4.184 J/^oC[/tex]

[tex]T_{final}[/tex] = final temperature = [tex]88.5^oC[/tex]

[tex]T_{initial}[/tex] = initial temperature = [tex]28^oC[/tex]

Now put all the given values in the above formula, we get:

[tex]q=220.9 g\times 4.184 J/g^oC\times (88.5-28)^oC[/tex]

[tex]q=55,916.86 J=55.916 kJ[/tex]  ( J = 0.001 kJ)

[tex]NaOH+HBr\rightarrow NaBr+H_2O[/tex]

Now we have to calculate the enthalpy change during the reaction.

[tex]\Delta H=-\frac{q}{n}[/tex]

where,

[tex]\Delta H[/tex] = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles NaOH = 1 mole

[tex]\Delta H=-\frac{55.916 kJ}{1 mol }=- 55.916 kJ/mol\approx 56.0 kJ/mol[/tex]

Therefore, the enthalpy change during the reaction is -56.0 kJ/mole.

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