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 work is partially correct, but there are a couple of mistakes in your calculations.
First, let's calculate the moles of KOH and H2SO4 correctly. The volume of KOH and H2SO4 solutions are the same, so the moles of KOH will also be 0.0272 mol, not 0.0272.
Next, when calculating the total volume, you correctly added the volumes of KOH and H2SO4, but you need to convert it to liters since the molar concentration is given in M (moles per liter). So, the total volume will be 54.4 mL = 0.0544 L.
Now, let's calculate the heat transferred (Q) using the formula Q = mcΔT. The mass of the solution is correct at 54.4 g, and the specific heat capacity of water is correct as 4.184 J/g°C. However, the temperature change (ΔT) is incorrect. It should be (30.17°C - 23.5°C) = 6.67°C, not (30.17°C - 23.5°C) = 6.67°C.
Let's correct the calculation:
Q = 54.4 g * 4.184 J/g°C * 6.67°C
Q = 1514.669 J
Now, to determine the value of ΔH (change in enthalpy) for the reaction, we need to divide the heat transferred (Q) by the moles of the limiting reactant, which is H2SO4. You correctly noted that for every 0.0136 mol of H2SO4, it releases 1516.74 J of heat. However, in your final calculation, you used the incorrect value of Q (1516.74 J).
Let's correct the calculation for ΔH:
ΔH = 1514.669 J / 0.0136 mol
ΔH = 111,372.19 J/mol
Finally, let's convert the units of ΔH to kilojoules (kJ):
ΔH = 111,372.19 J/mol = 111.37219 kJ/mol (rounded to three decimal places)
Therefore, the correct value for ΔH of the reaction is approximately -111.372 kJ/mol. Note that the negative sign indicates that the reaction is exothermic (heat is released).