you mix 45 ml of .20M KOH in calorimeter. The temperature of both reactions before mixing is 21.5 C. The Cp of the calorimeter was 36 J/K. If the final temperature of the mixture is 23.6 C, what is the enthalpy change per mole of water produced?

THE ENTHALPY CHANGE PER MOLE OF KOH IS 8400 Joules/ mole OF HEAT.

Explanation:

Heat = heat capacity * change in temperature

Heat capacity = 36 J/K

Temperature of the mixture before mixing = 21.5 C

Temperature of mixtire after mixing = 23.6 C

Calculate the change in temperature:

Change in temperature = 23.6 C - 21.5 C = 2.1 C

Heat = 36 * 2.1

Heat = 75.6 J of heat

In essence, 45 ml of 0.20 M of KOH produces 75.8 J of heat

The enthalpy change per mole of water:

It is important t obtain the number of moles involved in the reaction of 45 mL of 0.20 M of KOH

n = C V

n = 0.20 M * 45 *10^-3

n = 0.009 moles

Since number of moles = mass / molar mass

The mass of 45 ml of 0.20 M of KOH is then:

Molar mass = ( 39 + 16 + 1) g/mol = 56 g/mol

Mass = number of moles * molar mass

Mass = 0.009 * 56

Mass = 0.504 g

So therefore 0.504 g of KOH produces 75.6 J of heat

1 mole of KOH will produce x J of heat

1 mole of KOH = 56 g of KOH

0.504 g = 75.6 J

56 g = x J

x J = 56 * 75.6 / 0.504

x J = 8400 J / mole of KOH

3 1/4 %

explanation: