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: what do you think?
• 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
Your calculations seem correct so far. To calculate the change in enthalpy (delta H) of the reaction, you can use the heat released (Q) and the number of moles of sulfuric acid (H2SO4) reacted.
As you correctly calculated, the number of moles of H2SO4 is 0.0136 mol. The heat released (Q) is calculated using the formula Q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
In this case, the mass of the solution is 54.4 g (since the total volume is 54.4 mL and the density is 1.00 g/mL). The specific heat capacity of water (c) is 4.184 J/g°C. The change in temperature (ΔT) is 30.17°C - 23.5°C = 6.67°C.
Plugging in these values into the Q = mcΔT equation, you correctly calculated Q to be 1516.74 J.
Now, to calculate the change in enthalpy (delta H) of the reaction, you need to divide the heat released (Q) by the number of moles of H2SO4 reacted.
So, delta H = Q / n(H2SO4) = 1516.74 J / 0.0136 mol = -111.525 kJ/mol.
Rounding to the correct number of significant figures, the change in enthalpy (delta H) of the reaction is -112 kJ/mol. The negative sign indicates that the reaction is exothermic, meaning it releases energy to the surroundings.