When 24.30mL of .5 M H2SO4 is added to 24.30 of 1M KOH in a coffee cup calorimeter at 23.5 degrees C, the temp rises to 30.17 degrees C. Calculate delta H of this reaction.
(Assume that the total volume is the sum of the individual volumes and that density and specific heat capacity of the solution are the same as pure water d of water =1.00g/mL: c of water =4.184J/g*c))
answer in kJ/m H2SO4 ...................... please and thank you.
Is it -996 Kj/mol ?????????????????????
I couldn't get -996
To calculate the enthalpy change (ΔH) of the reaction, we can use the equation:
ΔH = q / n
where:
- ΔH is the enthalpy change
- q is the heat transferred
- n is the number of moles of the limiting reactant
First, we need to determine the number of moles of H2SO4 and KOH. To do this, we use the equation:
moles = concentration × volume
moles of H2SO4 = (0.5 M) × (0.02430 L)
moles of H2SO4 = 0.01215 moles
moles of KOH = (1 M) × (0.02430 L)
moles of KOH = 0.02430 moles
Next, we need to calculate the heat transferred (q) using the equation:
q = m × c × ΔT
where:
- q is the heat transferred
- m is the mass of the solution (which we can calculate using the density)
- c is the specific heat capacity of the solution (given as 4.184 J/g°C)
- ΔT is the change in temperature
The mass (m) of the solution can be calculated as follows:
mass = volume × density
mass = (0.02430 L + 0.02430 L) × (1.00 g/mL)
mass = 0.04860 kg
ΔT = final temperature - initial temperature
ΔT = 30.17°C - 23.5°C
ΔT = 6.67°C
Now, we can calculate the heat transferred (q):
q = (0.04860 kg) × (4.184 J/g°C) × (6.67°C)
q = 1.7188 kJ
Finally, we can calculate the enthalpy change (ΔH):
ΔH = q / n
ΔH = (1.7188 kJ) / 0.01215 moles
ΔH = -141.57 kJ/mol
Therefore, the ΔH of the reaction is approximately -141.57 kJ/mol.