Suppose H2A reacts with ZOH according to this reaction:
H2A(aq) + 2ZOH(aq) --> 2H2O(l) + Z2A(s)
When 100.0 mL of 0.175 M H2A and 100.0 mL of 0.225 M ZOH, both initially at 20.6 oC, are mixed in a coffee cup calorimeter, the temperature rises to 23.9 oC and solid Z2A forms.
Δ H rxn per mol of Z2A produced.
Oxygen and Hydrogen are the usual elements.
The atomic mass of A is 15.9854 g/mol.
The atomic mass of Z is 35.5680 g/mol
Assume the density of the solution is 1.0 g/mL and the heat capacity is that of water, 4.184 J/goC.
I suppose you assume that the heat capacity of the solid, Z2A, is ignored.
H2A(aq) + 2ZOH(aq) --> 2H2O(l) + Z2A(s)
mols H2A initially = M x L = 0.175 M x 0.100 L = 0.0175
mols ZOH initially = 0.225 x 0.100 = 0.0225
The limiting reagent (LR) is ZOH so 0.0225 of ZOH will react with 1/2 x 0.0175 = 0.00875 mols H2A leaving 0.0175 - 0.00875 = ?
oops! I hit the wrong button and posted before I finished the problem.
I suppose you assume that the heat capacity of the solid, Z2A, is ignored.
H2A(aq) + 2ZOH(aq) --> 2H2O(l) + Z2A(s)
mols H2A initially = M x L = 0.175 M x 0.100 L = 0.0175
mols ZOH initially = 0.225 x 0.100 = 0.0225
The limiting reagent (LR) is ZOH so 0.0225 of ZOH will react with 1/2 x 0.0175 = 0.00875 mols H2A leaving 0.0175 - 0.00875 = ?
q = mass x specific heat x delta T
mass solution = 200 mL H2O with a density of 1.0 g/mL = 200 g and we ignore the ppt of Z2A so
q = 200 g x 4.184 J/g*C x ( 23.9 - 20.6) = about 2761 J but you need to go through these calculations again to get better numbers.
That's 2761 J from 0.0225 mols ZOH to produce 0.0225 mols H2O (or 1/2 that of Z2A).
2761 J for 0.0225 mols H2O = 2761 x 1/0.0225 = 122, 711 J or 122.711 kJ/2 mols H2O = about 61 kJ/mol H2O or per mol Z2A produced. You will get a slightly different value if you recalculate from the beginning.
To find the enthalpy change per mol of Z2A produced (ΔH_rxn), we can use the equation:
ΔH_rxn = q / n
where q is the heat absorbed or released by the reaction and n is the number of moles of Z2A produced.
First, we need to calculate the heat absorbed or released by the reaction (q). We can use the formula:
q = m * C * ΔT
where m is the mass of the solution, C is the heat capacity, and ΔT is the change in temperature.
To calculate the mass of the solution, we add the masses of the H2A and ZOH solutions. Since their densities are both 1.0 g/mL, their masses are equal to their volumes:
Mass of H2A solution = Volume of H2A solution * Density of the solution
= 100.0 mL * 1.0 g/mL
= 100.0 g
Mass of ZOH solution = Volume of ZOH solution * Density of the solution
= 100.0 mL * 1.0 g/mL
= 100.0 g
Total mass of the solution = Mass of H2A solution + Mass of ZOH solution
= 100.0 g + 100.0 g
= 200.0 g
Next, we calculate the change in temperature (ΔT):
ΔT = Final temperature - Initial temperature
= 23.9 oC - 20.6 oC
= 3.3 oC
Now, we can calculate q:
q = m * C * ΔT
= 200.0 g * 4.184 J/goC * 3.3 oC
= 2739.84 J
Next, we need to calculate the number of moles of Z2A produced. From the balanced chemical equation, we can see that for every 2 moles of ZOH, 1 mole of Z2A is produced. Given the molarity of the ZOH solution, we can calculate the number of moles:
Number of moles of ZOH = Molarity * Volume
= 0.225 M * 100.0 mL
= 0.0225 mol
Moles of Z2A produced = 0.0225 mol * (1 mol Z2A / 2 mol ZOH)
= 0.01125 mol
Now, we can calculate the enthalpy change per mol of Z2A produced (ΔH_rxn):
ΔH_rxn = q / n
= 2739.84 J / 0.01125 mol
= 243,088 J/mol
However, the question asked for the answer in kJ/mol. Therefore, we need to convert the answer to kilojoules:
ΔH_rxn = 243,088 J/mol / 1000
= 243.088 kJ/mol
So, the enthalpy change per mol of Z2A produced is 243.088 kJ/mol.