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.
