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, sir
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, sIr
This value is for 2 mols H2O so you need to to do -112 kJ/mol x 1/2 = about 56 kJ/mol.
Is that a -56 ?
Your work is correct. To calculate the delta H of the reaction, you first need to calculate the heat transferred, which is given by the equation Q=mcΔT, where Q is the heat transferred, m is the mass of the solution, c is the specific heat capacity, and ΔT is the change in temperature.
In your calculations, you correctly determined the number of moles of KOH and H2SO4, which are 0.0272 mol and 0.0136 mol, respectively. The total volume of the solution is 54.4 mL (the sum of the individual volumes), and since the density of water is 1.00 g/mL, the mass of the solution is 54.4 g.
Using the values you calculated, you can plug them into the equation Q=mcΔT. The mass of the solution is 54.4 g, the specific heat capacity of water is 4.184 J/g°C, and the change in temperature is 30.17°C - 23.5°C = 6.67°C.
Plugging in these values, you correctly calculated Q to be 1516.74 J.
Since the heat is lost from the system to the surroundings, the delta H of the reaction is negative.
To find the delta H for one mole of H2SO4, you divide the heat transferred (Q) by the number of moles of H2SO4 (0.0136 mol).
So, delta H = 1516.74 J / 0.0136 mol = -111.525 kJ/mol.
Rounding to three significant figures, the delta H is approximately -112 kJ/mol.