You mix 45 ml of .20M HCl with 45 ml of .20M KOH in a calorimeter. The temperature of both reactions before mixing is 21.5oC. The temperature rises for the mixture to 22.7oC after mixing. What is the enthalpy change per mole of water produced?
KOH + HCl ==>H2O + KCl
mols HCl = M x L = ?
mols KOH = M x L = ?
mols HCl = mols KOH so the reaction is exactly neutralized and mols H2O prduced is 0.009.
q rxn = mass solution x specific heat solution x delta T
The problem doesn't say what the density is; I assume we are to assume it is 1.00 g/mL. There are 90 mL when mixed so g H2O is 90 grams.
dq = 90 x 4.184 x delta T = approximately 450 but you need a more accurate answer. So
q rxn = 450 J and that is for 0.009
delta H/mol = q/mol = approx 450/0.009 = ? and it is exothermic so the sign of delta H is negative. Most change this to delta H = -??? kJ/mol.
To find the enthalpy change per mole of water produced, we need to use the equation:
ΔH = q / n
where ΔH is the enthalpy change, q is the heat released or absorbed by the system, and n is the number of moles of the substance.
First, we need to find the heat released or absorbed by the system. We can use the equation:
q = mCΔT
Where q is the heat released or absorbed, m is the mass of the solution, C is the specific heat capacity of the solution, and ΔT is the change in temperature.
Since we have equal volumes of HCl and KOH (45 mL each), we can assume the total volume of the solution is 90 mL. The density of water is around 1 g/mL, so the mass of the solution can be estimated to be 90 g.
By assuming the specific heat capacity of the solution is the same as water (4.18 J/g°C), we can calculate the heat released or absorbed:
q = (90 g)(4.18 J/g°C)(22.7 - 21.5)°C
q = 372.6 J
Now, we need to calculate the number of moles of water produced. Since HCl and KOH react in a 1:1 ratio, the number of moles of water produced is equal to the number of moles of limiting reactant used.
The volume of HCl used is 45 mL, which is equal to 0.045 L. By using the equation:
moles = concentration × volume
we find:
moles of HCl = (0.20 mol/L)(0.045 L)
moles of HCl = 0.009 mol
Since HCl and KOH react in a 1:1 ratio, the number of moles of water produced is also 0.009 mol.
Now we can calculate the enthalpy change per mole of water produced:
ΔH = q / n
ΔH = 372.6 J / 0.009 mol
ΔH = 41,400 J/mol
Therefore, the enthalpy change per mole of water produced is 41,400 J/mol.