Question

An experiment is run in 1000.00mL solution, where 0.500mol of a metal are reacted with an excess of acid. If the initial temperature of the reaction mixture was 24.00 degree C and the reaction had a delta H of 458.5 kJ/mol, What is the final temperature of the reaction solution? Assume that there is the final temperature of the reaction solution? Assume that there are no competing reactions, the solution's density and specific heat are identical to those of water (i.e. 1.00g/mLand 4.184 J/g degree C, respectively). Was the reaction above exothermic or endothermic? Now suppose that instead of just one reaction, there was also a competing reaction going on in the same vessel at the same time. If the competing reaction had a delta H of 79kJ/mol and consumed exactly 10% of the starting material, what would the final temperature of the solution be? Assume that there were no further competing reactions, and that the products do not change the properties of the solution (i.e. same density and specific heat as water).

Answer 1

Ans. Total heat released during reaction of metal with acid =

                                    Moles of metal x Molar enthalpy of reaction of metal with acid

                                    = 0.500 mol x (- 485.5 kJ/mol)

                                    = - 242.75 kJ

                                    = - 242750 J                                                  ; [1 kJ = 1000 J]

Note that the (-ve) sign indicates that the reaction of metal with acid releases energy.

It’s assumed that all the energy released during reaction of metal with acid is absorbed by the solution.

Now, using              q = m x s x dT                      - equation 1   

Where,

q = heat gained (or lost)

m = mass of water in gram

s = specific heat, [of water = [ for water, s = 4.184 J g^-10C^-1]

dT = change in temperature = Final temperature, T2 – Initial temperature, T1

Putting the values in equation 1-

            242750 J = 1000 g x (4.184 J g^-10C^-1) x (T2 – 24.00⁰C)

            Or, 242750 J / (4184 J C^-1) = T2 – 24.00⁰C

            Or, 58.02⁰C + 24.00⁰C = T2

            Or, T2 = 82.02⁰C

Therefore, final temperature = 82.02⁰C

# The reaction described above, between metal and acid, is exothermic because it releases heat as indicated by (-ve) value of dH.

#2. Moles of acid consumed by competing reaction = 10% of 0.500 mol

                                                                                                = 0.050 mol

Remaining moles of metal to react with acid =

Initial moles of metal – Moles of metal consumed in competing reaction

            = 0.500 mol – 0.050 mol

            = 0.450 mol

Total heat released during reaction of metal with acid =

                                    Moles of metal x Molar enthalpy of reaction of metal with acid

                                    = 0.450 mol x (- 485.5 kJ/mol)

                                    = - 218.475 kJ

                                    = - 218475 J

Assumption: The stoichiometry of competing reaction with metal is not mentioned. So, it’s assumed that the competing reactions consumes 1 mol of metal per mol of the reaction.

So, moles of active reactant in competing reaction consumed up = Moles of metal consumed up in it = 0.05 mol.

Amount of heat absorbed by competing reaction = Molar enthalpy of competing reaction x Moles of active reactant consumed up

                                                = (79 kJ/mol) x 0.05 mol

                                                = 3.95 kJ

                                                = 3950 J

Amount of heat available to be absorbed by the solution = Total heat released during metal-acid reaction – Heat absorbed by competing reaction

                                                = 218475 J – 3950 J

                                                = 214525 J

Therefore, due to competing reaction, there is only 214525 J available to be absorbed by the solution.

Putting the values in equation 1-

            218475 J = 1000 g x (4.184 J g^-10C^-1) x (T2 – 24.00⁰C)

            Or, 218475 J / (4184 J C^-1) = T2 – 24.00⁰C

            Or, 52.22⁰C + 24.00⁰C = T2

            Or, T2 = 76.22⁰C

Therefore, final temperature = 76.22⁰C