Question

If 30.0 ml of 0.150 m aqueous sodium hydroxide is mixed with 30.0 ml of 0.150 m aqueous hydrochloric acid in a calorimeter at an initial temperature of 25.0 degrees celsius, what is the enthalpy change of this reaction if the final temperature reached in the calorimeter is 27.5 degrees celsius? naoh + hcl nacl + h2o

Answer 1

Answer : As given in the question the concentration of acid and base are in equimolar proportion so,

the simple net ionic reaction that the system is undergoing can be given as,
$H^{+} _{(aq.)} + OH^{-} _{(aq.)} ----\ \textgreater \ H_{2} O + heat$

we can now calculate the total number of moles reacting in the species as,

(30 mL) X (1L/1000ML) X 0.150 moles = $4.5 X 10 ^{-3}$ moles.

Now, we know the specific heat of water as 4.186 J/g °C.

Now considering the total volume of the product as 60 mL or 60 g (as it is water it can be g or mL) hence, ΔC is 2.5 C.

Now converting these into joules = 4.186 J/g °C X 2.5 X 60 = 627.9 Joules.

On dividing this we get 627.9 Joules / $4.5 X 10 ^{-3}$ moles = 140 KiloJoules/mole.

So the answer is 140 KJ/M.