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student0724
24.07.2020 •
Chemistry
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?
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Ответ:
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
Ответ: