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 calculation for the heat released by the reaction is correct. However, there is a small error in the calculation for the value of delta H. Let's go through the steps and I'll show you how to calculate it correctly.
Step 1: Calculate the moles of KOH and H2SO4
n(KOH) = 27.20 mL * 1 M = 0.0272 mol
n(H2SO4) = 27.20 mL * 0.5 M = 0.0136 mol
Step 2: Calculate the total mass of the solution
Total volume = 27.20 mL + 27.20 mL = 54.40 mL
Mass of solution = total volume * density of water
Mass of solution = 54.40 g/mL * 1.00 g/mL = 54.40 g
Step 3: Calculate the heat released (Q)
Q = mass of solution * specific heat capacity of water * change in temperature
Q = 54.40 g * 4.184 J/g°C * (30.17°C - 23.5°C)
Q = 1586.59 J
Step 4: Calculate delta H
The heat released by 0.0136 mol of H2SO4 is 1586.59 J, so for 1 mole of H2SO4, the heat released would be:
Delta H = Q / n(H2SO4)
Delta H = 1586.59J / 0.0136 mol
Delta H = -116,570 J/mol
Delta H = -116.6 kJ/mol (rounded to one decimal place)
Therefore, the correct value for delta H of the reaction is approximately -116.6 kJ/mol.