Heat of reaction calculations (1 Viewer)

[hydrobiont]

Q1. When 25ml of a solution at 23.2 degrees celcius containing 5.00*10-3 mol sodium hydroxide was mixed with 50ml of a solution also at 23.2 degrees celcius that contained 6.00*10-3 mol hydrochloric acid in a light plastic beaker, the final temperature of the mixture was 24.1 degrees celcius. Calculate the heat released and hence the enthalpy change per mole for the neutralisation reaction. Take the density and specific heat capacity of all solutions involved as 1.00g/ml and 4.2 J K-1 g-1 respectively, assume that the container had negligible heat capacity and that heat losses to the surroundings were negligible.


Q2. 50ml of a 0.28mol/L solution of ammonia at 19.6 degrees celcius was added to 100ml of a 0.14mol/L solution of nitric acid also at 19.6 degrees celcius. The final temperature was 20.7 degrees celcius. Write a net ionic equation for the reaction that occured. Making the same assumptions and using the same values for density and heat capacity as in Q1, calculate the enthalpy change for this reaction.



Correct answers:
Q1 57 KJ/mol ; -57 KJ/mol

Q2 enthalpy change: -50 KJ/mol


Does anyone know how to do the above two Qs.
I tried but couldn't get the correct answers.

 

[Dreamerish*~]

Q1. When 25ml of a solution at 23.2 degrees celcius containing 5.00*10-3 mol sodium hydroxide was mixed with 50ml of a solution also at 23.2 degrees celcius that contained 6.00*10-3 mol hydrochloric acid in a light plastic beaker, the final temperature of the mixture was 24.1 degrees celcius. Calculate the heat released and hence the enthalpy change per mole for the neutralisation reaction. Take the density and specific heat capacity of all solutions involved as 1.00g/ml and 4.2 J K-1 g-1 respectively, assume that the container had negligible heat capacity and that heat losses to the surroundings were negligible.

NaOH + HCl → NaCl + H₂O

n_NaOH = 0.025 x (5.00 x 10^-3) = 0.000125
n_HCl = 0.05 x (6.00 x 10^-3) = 0.0003

.: NaOH is in excess, and that in this reaction, 0.000125 moles of NaOH will react with 0.00125 moles of HCl.

ΔH = MCΔT

ΔH = 75 x 4.2 x 0.9 = 283.5 J

So 283.5 J was released from every 0.000125 moles. So the enthalpy change per mole is 283.5/0.000125 = 2268000 J = 2268 kJ

In 1 gram of NaOH there are 1/40 moles. 2268 x 1/40 = 56.7

So ΔH = 56.7 kJ per gram, not per mole. The question asked for per mole but the answer was given for per gram. Either you copied the question incorrectly or whoever wrote the answer wasn't paying attention.

Q2. 50ml of a 0.28mol/L solution of ammonia at 19.6 degrees celcius was added to 100ml of a 0.14mol/L solution of nitric acid also at 19.6 degrees celcius. The final temperature was 20.7 degrees celcius. Write a net ionic equation for the reaction that occured. Making the same assumptions and using the same values for density and heat capacity as in Q1, calculate the enthalpy change for this reaction.

NH₃ + HNO₃ → NH₄⁺ + NO₃^-

n_NH3 = 0.05 x 0.28 = 0.014
n_HNO3 = 0.1 x 0.14 = 0.014

.: 0.014 moles of NH₃ is reacting with 0.014 moles of HNO₃.

ΔH = MCΔT

ΔH = 150 x 4.2 x 1.1 = 693 J

For one mole, ΔH = 693/0.014 = 49500 J = 49.5 kJ.

Since the reaction is exothermic, ΔH = -49.5 kJ.

 

[hydrobiont]

thank you for your help