Question

A 100.0 mL of 0.500 M HBr at 20.29 oC is added to 100.0 mL of 0.500 M KOH (also at 20.29 oC). After mixing, the temperature rises to 23.65 oC. Calculate the heat of this reaction. [assuming the density and specific heat of HBr and KOH is the same as water, 1.0 g/mL; 4.18 J/g oC, and the volume of the solution is additive].

Answer 1

Answer : The heat of this reaction will be, 2.81 kJ

Explanation :

First we have to calculate the mass of water.

`\text{Mass of water}=\text{Density of water}* \text{Volume of water}`

Given:

Density of water = 1.0 g/mL

Volume of water = 100.0 mL + 100.0 mL = 200.0 mL

`\text{Mass of water}=1.0g/mL* 200.0mL`

`\text{Mass of water}=200.0g`

Now we have to calculate the heat of this reaction.

Formula used :

`q=m* c* (T_(final)-T_(initial))`

where,

q = heat = ?

m = mass of water = 200.0 g

c = specific heat of water =
`4.18J/g^oC`

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

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

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

`q=200.0g* 4.18J/g^oC* (23.65-20.29)^oC`

`q=2808.96J=2.81kJ`

Thus, the heat of this reaction will be, 2.81 kJ