When 25.0 ml of 0.500M
HCI is added to 25.0ml of 0.500M KOH in a coffee cup calorimeter at 23.50C, the temperature rises to 30.17C. Calculate delta H of this reaction ( assume density of solution = 1g/ml)
NaOH + HCl ==> NaCl + H2O
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
mols H2O formed is M x L = 0.0125
dH = q/0.0125 J/mol
To calculate the enthalpy change (ΔH) of a reaction using calorimetry, you can use the equation:
ΔH = q / n
where ΔH is the enthalpy change, q is the heat absorbed or released by the reaction, and n is the number of moles of the limiting reactant. In this case, we will assume that the limiting reactant is either HCl or KOH.
To find q, you can use the equation:
q = m * c * ΔT
where q is the heat absorbed or released, m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature.
First, let's find the mass of the solution. Since the density of the solution is given as 1 g/mL, the mass of the solution is the same as its volume. Therefore, the mass of the solution is:
mass = 25.0 mL + 25.0 mL = 50.0 g
Next, we need to find the change in temperature, ΔT:
ΔT = final temperature - initial temperature
= 30.17°C - 23.50°C
= +6.67°C
Now, since we are considering the reaction between HCl and KOH, we need to find the number of moles of each reactant using the equation:
moles = concentration * volume
For HCl:
moles of HCl = 0.500 M * 25.0 mL / 1000 mL/L
= 0.0125 moles
For KOH:
moles of KOH = 0.500 M * 25.0 mL / 1000 mL/L
= 0.0125 moles
Since both reactants have the same number of moles and concentration, it means they react in a 1:1 ratio.
Now, we can calculate q using the equation mentioned earlier:
q = m * c * ΔT
= 50.0 g * 4.18 J/g°C * 6.67°C
= 13985 J
Finally, we can calculate ΔH using the equation:
ΔH = q / n
= 13985 J / 0.0125 moles
= 1118800 J/mol
Therefore, the enthalpy change (ΔH) of this reaction is 1118800 J/mol.