H2 Chemistry Thermochemistry
-
1. When 2.76g (0.02mol) of potassium carbonate was added to 30cm^3 of approximately 2moldm^-3 hydrochloric acid, the temperature rose by 5.2drg celcius.
i) write an equation for the reaction
ii) calculate the enthalpy change of this reaction per mole of potassium carbonate. Assume that the specific heat capacities of all solutions have a density of 1.0gcm^-3. Final Answer: -32.6kJmol^-1
iii) Explain why the hydrochloric acid need only be approximately 2moldm^-3.
this is my equation: K2CO3(s) + 2HCl (aq) -> CO2(g) + H20(l) + 2KCl (aq, and the limiting reagent is K2CO3 right?
but i dunno how to answer part ii) and iii) at all...
-
Originally posted by qdtimes2:
1. When 2.76g (0.02mol) of potassium carbonate was added to 30cm^3 of approximately 2moldm^-3 hydrochloric acid, the temperature rose by 5.2drg celcius.
i) write an equation for the reaction
ii) calculate the enthalpy change of this reaction per mole of potassium carbonate. Assume that the specific heat capacities of all solutions have a density of 1.0gcm^-3. Final Answer: -32.6kJmol^-1
iii) Explain why the hydrochloric acid need only be approximately 2moldm^-3.
this is my equation: K2CO3(s) + 2HCl (aq) -> CO2(g) + H20(l) + 2KCl (aq, and the limiting reagent is K2CO3 right?
but i dunno how to answer part ii) and iii) at all...
(i) CO3 2- + 2H+ ---> H2CO3 ---> CO2 + H2O
(ii) Heat transferred = m x c x delta T = 30 x 4.18 x 5.2 = 652.08
Since temperature rose, reaction is exothermic, ie. enthalpy change per 0.02 moles of CO3 2- = -652.08 J
enthalpy change per mole of CO3 2- = -652.08 / 0.02 = -32,604 J = -32.6 kJ/mol
(iii) Based on stochiometry of the limiting reactant CO3 2-, we only need 0.04 mol of H+, implying we only need 1.33 mol/dm3 of HCl(aq). To ensure that CO3 2- is limiting (since our objective is to determine the enthalpy change of reaction per mole of CO3 2-), hence our molarity of HCl should be slightly above 1.33 mol/dm3, for instance 2 mol/dm3.
-
hmm...why don't we need to consider the mass of potassium carbonate when calculating the heat involved?
-
Originally posted by qdtimes2:
hmm...why don't we need to consider the mass of potassium carbonate when calculating the heat involved?
Because the potassium carbonate isn't the species which absorbed the heat. -
hmm....then why is it that we take into account of the total mass, in let's say a reaction between hydrochloric acid and sodium hydroxide?
Example of the qn in my notes:
50cm^3 of 1moldm^-3 HCl were added to 50cm^3 of 0.65moldm^-3 NaOH in a plastic beaker. The temperature of the resulting solution changed from 21.0 to 25.5 drg celcius. Assuming that it takes 4.2J to increase the temperature of 1cm^3 of solution by 1.0 drg celcius, calculate the enthalpy change of neutralisation per mol of water.Answer: Q = (50+50) x 4.2 x (25.5 - 21.0) = 1.89 kJ
-
Originally posted by qdtimes2:
hmm....then why is it that we take into account of the total mass, in let's say a reaction between hydrochloric acid and sodium hydroxide?
Example of the qn in my notes:
50cm^3 of 1moldm^-3 HCl were added to 50cm^3 of 0.65moldm^-3 NaOH in a plastic beaker. The temperature of the resulting solution changed from 21.0 to 25.5 drg celcius. Assuming that it takes 4.2J to increase the temperature of 1cm^3 of solution by 1.0 drg celcius, calculate the enthalpy change of neutralisation per mol of water.Answer: Q = (50+50) x 4.2 x (25.5 - 21.0) = 1.89 kJ
Because the reactants in this particular question are aqueous, meaning the water from these aqueous reactants must be considered to have absorbed the heat evolved. -
i see... thanks a lot! :)
-
Originally posted by qdtimes2:
i see... thanks a lot! :)
You're welcome!
You may wish to attempt this releated question on Energetics, but note that I will *not* provide the full worked solution, but only the final answers. Have fun!
--------------------------------------
Liquid butane is commonly found in canisters used for portable gas stoves during camping trips. The standard enthalpy change of combustion of butane is -2887 kJ/mol. A camper found a left-over canister of butane and estimates that there was 60 g of liquid butane left.
a) Calculate the mass of water, at r.t.p, that he can boil before he uses up the butane.
b) Using a Hess Law energy cycle diagram, and given that the standard enthalpies of formation of CO(g) and CO2(g) are -111kJ/mol and -394kJ/mol respectively, calculate the standard enthalpy change of reaction for the incomplete combustion of butane to produce water and carbon monoxide only.
Answers :
a) 9.49 kg
b) -1745 kJ/mol