Question

2. When 2.0 g of NaOH were dissolved in 53.0 g water in a calorimeter at 24.0°C, the temperature of the solution went up to 33.7°C. Why? Is this solution reaction exothermic? а. b. Calculate qH2o using Eq. 1 Find AH for the reaction as it occurred in the calorimeter (Eq. 5). с. Find AH for the solution of 1.00 g NaOH in water d. J/g Find AH for the solution of one mole NaOH in water. е. J/mol Given that NaOH exists as Na' and OH ions in solution, write the equation for the reaction that occurs f. when NaOH is dissolved in water. g. Using enthalpies of formation as given in thermodynamic tables, calculate AH for the reaction in Part f compare your answer with the result you obtained in Part e and b0

Answer 1

1- Here given when NaOH is added to water, the temperature increases from 24.0^oC to 33.7^oC. That means there must be a reaction between NaOH and water takes place, in which heat is released and this heat increased the temperature of the solution. So as heat is released during the reaction, this reaction is Exothermic in nature.

2- Now Q_H2O is the amount of heat absorbed by the water where its temperature increased from 24.0^oC to 33.7^oC. This amount of heat absorbed by water is calculated by the formula

Q_H2O = m * C * ΔT

where m = mass of water taken

C = specific heat of water

ΔT = change in temperature = final temp - initial temp

Now putting the given values-

Q_H2O = m * C * ΔT

= 53 g * 4.186 J/g^oC * (33.7^oC - 24.0^oC)

= 53 g * 4.186 J/g^oC * (9.7‬^oC)

= 2152 J

3- Now this heat that is absorbed by water is actually released from the reaction. So the heat of reaction = ΔH = -2152 J

Here the negative sign indicates the reaction is Exothermic in nature.

4-

In the above question, we get heat of reaction for 2 g od reaction = -2152 J

Now the heat of reaction per 1 g of NaOH = -2152 J/ 2g

= -1076 J/g

5- Now we know mass of 1 mole NaOH = 40 g/mole

Then moles of NaOH in 2 g NaOH = mass / molar mass

= 2 g / 40 g/mole

= 0.05 moles

Then

heat of reaction per 1 mole of NaOH = -2152 J/ 0.05 moles

= -43040‬ J/ mole