20cm3 of 0.500moldm-3 H2SO4 was consumed to neutralize 50cm3 of 0.400moldm-3 NaOH. The temperature of the system is increased by 36 degree C. The heat capacity of the calorie meter was 39 JK-1, and the heat capacityof the solution is 4.2JK-1. Find the enthalpy of neutralization for the reaction.
Well, let me calculate that for you!
First, we need to calculate the heat released during the reaction using the formula:
Q = m × c × ΔT
Where:
Q = Heat released (in Joules)
m = Mass of the solution (in grams)
c = Specific heat capacity of the solution (in J/g°C)
ΔT = Change in temperature (in °C)
We need to calculate the mass of the solution first. Since the density of water is 1 g/cm³, the mass of the solution is equal to its volume.
The volume of the solution = 50 cm³
Thus, the mass of the solution = 50 g.
Now, let's calculate the heat released using the given values:
c = 4.2 J/g°C
ΔT = 36°C
Q = (50 g) × (4.2 J/g°C) × (36°C)
Q = 7560 J
Since the heat capacity of the solution is 4.2 J/K, we need to convert the heat released to heat evolved in the reaction. To do that, we divide the heat released by the heat capacity:
Heat evolved = Q / Heat Capacity of the Solution
Heat evolved = 7560 J / 4.2 J/K
Heat evolved ≈ 1800 K
Finally, we divide the heat evolved by the number of moles of H2SO4 used to neutralize the NaOH to find the enthalpy of neutralization (ΔH).
ΔH = Heat evolved / (Number of Moles of Acid)
ΔH = 1800 K / (0.500 mol/dm³ × 0.020 dm³)
ΔH = 1800 K / 0.010 mol
So, the enthalpy of neutralization for the reaction is approximately 180,000 J/mol.