Part A - In one experiment, a student placed 50.0 mL of 1.00 M acetic acid (HC 2
ID: 511868 • Letter: P
Question
Part A -
In one experiment, a student placed 50.0 mL of 1.00 M acetic acid (HC2H3O2) at 25.5 °C in a coffee cup calorimeter. To this was added 50.0 mL of 1.50 M NaOH solution. The mixture was stirred, and the temperature quickly increased by +6.6 °C.
a) The net ionic equation is: (Include physical states in your answer. Use minimal integer numbers for coefficient input.)
b) What is the Hº for this reaction?
kJ·mol-1
Part B -
What is the thermochemical equation for the formation of 2.5 mol of NH3(g) from H2(g) and N2(g)?
First, enter the chemical equation:
Next, determine H (kJ)
Explanation / Answer
Net ionic equation: HC2H3O2(aq) + OH-(aq) --> H2O(l) + C2H3O2-(aq)
Molarity of HC2H3O2 = 1 M
Volume of HC2H3O2 = 50 mL = 0.05 liters
Molarity of NaOH = 1.50 M
Volume of NaOH = 50 mL = 0.05 liters
moles HC2H3O2 = Molarity x Volume = 1.00 x 0.050 = 0.05
moles NaOH = Molarity x Volume = 1.50 x 0.050 = 0.0750
Assumming that specific heat of solution = specific heat H2O = 4.18 J/g. 0C
and density of solution = density of water = 1 gm/mL
Total volume of solution = 50 mL of HC2H3O2 + 50 mL of NaOH = 100mL
Mass of solution = Density of solution x Total volume of solution = 1 x 100 = 100 gm
HC2H3O2(aq) + NaOH(aq) = NaC2H3O2(aq) + H2O(l)
q = mass of solution x specific heat solns x (Change in temperature of solution)
q = 100g x 4.18 x (6.6) = 2758.8 J
H0 in J/mol= q/0.05 mol HC2H3O2 = 55176 J/mol of HC2H3O2 = 55.176 kJ/mol
Part B
Chemical equation for one mole of NH3 : (1/2)N2(g) + (3/2)H2(g) --------> NH3(g)
Heat of formation of NH3 (298 K) = -46.1 kJ/mol
Heat of formation of N2 (298 K) = 0kJ/mol
Heat of formation of H2 (298 K) = 0 kJ/mol
Hr0 (298 K) = Heat of formation of NH3 (298 K) - [(1/2) x Heat of formation of N2 (298 K)+ (3/2) x Heat of formation of H2 (298 K)] =-46.1 - [ (1/2)x0 + (3/2) x 0] = -46.1 kJ
Thermochemical equation for one mole of NH3 : (1/2)N2(g)+ (3/2)H2(g) --------> NH3(g) Hr0 = -46.1 kJ
This is for 1 mole of NH3, for 2.5 or 5/2 moles of NH3 multiplying thermochemical equation for one mole by (5/2)
(5/2) x (1/2) N2 + (5/2) x (3/2) H2 -------> (5/2) NH3 H = 5/2 Hr0
(5/4) N2 + (15/4)H2 ------> (5/2) NH3 H = (5/2) x (- 46.1) = -115.25 kJ
Thermochemical equation for 2.5 mol of NH3 : (5/4) N2(g) + (15/4)H2 (g) ------> (5/2) NH3 (g) H = -115.25 kJ