When 27.20mL of .5 M H2SO4 is added to 27.20 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))
my work: correct or no??
• 2KOH + H2SO4→ 2H2O + K2SO4
n(KOH) = 0.0272 * 1 = 0.0272mol
n(H2SO4) =0.0272 * 0.5 = 0.0136 mol
total volume = 27.2 + 27.2 = 54.4mL
mass of solution = 54.4mL * 1g/mL =54.4g
Q = mcΔT
= 54.4g * 4.18 J/g°C * (30.17 - 23.5)°C
= 1516.74J
The heat is lost from system to surrounding, so ΔH is-ve
For every 0.0136 mol of H2SO4 react, it releases 1516.74J of heat
For 1 mole of H2SO4,
ΔH= 1516.74/0.0136
= -111.525 kJ / mol
= -112 kJ/ mol.......thank you
Your calculations appear to be correct. Let's go through the process step by step:
1. First, we need to calculate the number of moles of KOH and H2SO4. We are given that the volume of each solution is 27.20 mL, and their concentrations are 1 M and 0.5 M, respectively.
n(KOH) = volume(KOH) * concentration(KOH) = 0.0272 L * 1 M = 0.0272 mol
n(H2SO4) = volume(H2SO4) * concentration(H2SO4) = 0.0272 L * 0.5 M = 0.0136 mol
2. Next, we determine the total volume of the solution. The total volume is the sum of the individual volumes of KOH and H2SO4.
total volume = volume(KOH) + volume(H2SO4) = 27.20 mL + 27.20 mL = 54.4 mL = 0.0544 L
3. To calculate the mass of the solution, we use the density of water (since the density and specific heat capacity of the solution are assumed to be the same as pure water).
mass of solution = volume of solution * density of water = 0.0544 L * 1 g/mL = 54.4 g
4. Now we can use the heat equation, Q = mcΔT, to calculate the heat energy transferred. The specific heat capacity of water is given as 4.184 J/g°C.
Q = mass of solution * specific heat capacity of water * change in temperature
= 54.4 g * 4.184 J/g°C * (30.17 - 23.5)°C
≈ 1516.74 J
Note: The negative sign is due to the fact that the heat is lost from the system to the surroundings.
5. Finally, we need to calculate ΔH, which represents the enthalpy change of the reaction per mole of H2SO4. Since for every 0.0136 mol of H2SO4 reacted, 1516.74 J of heat is released, we can use this ratio to calculate ΔH for 1 mole of H2SO4.
ΔH = (heat released per mole of H2SO4) / (number of moles of H2SO4)
= 1516.74 J / 0.0136 mol
≈ -111.525 kJ/mol
≈ -112 kJ/mol
Therefore, the ΔH of the reaction is approximately -112 kJ/mol.
Well done on your calculations! If you have any more questions, feel free to ask.