Question

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 =

Answer 1

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

Step-by-step 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.

`q=mc* (T_(final)-T_(initial))`

where,

m = mass of solution

q = heat gained = ?

c = specific heat =
`4.184 J/^oC`

`T_(final)` = final temperature =
`88.5^oC`

`T_(initial)` = initial temperature =
`28^oC`

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

`q=220.9 g* 4.184 J/g^oC* (88.5-28)^oC`

`q=55,916.86 J=55.916 kJ` ( J = 0.001 kJ)

`NaOH+HBr\rightarrow NaBr+H_2O`

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

`\Delta H=-(q)/(n)`

where,

`\Delta H` = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles NaOH = 1 mole

`\Delta H=-(55.916 kJ)/(1 mol )=- 55.916 kJ/mol\approx 56.0 kJ/mol`

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