250cm3 of 0.400moldm-3 NaOH and 250cm3 of 0.400moldm-3 HCl is allowed to react in a calorie meter. If the initial temperature of the system is 17.05 degree C and the final temperature is 19.55 degree C,Find the enthalpy of neutralization for the reaction. The mass of the calorie meter is 500g. Specific heat capacities of the calorie meter and the solutions are 400JKg-1K-1 and 4200JKg-1K-1 respectively
q = [mass solns x specific heat solns x (Tfinal-Tinitial)] + [mass Cal x specific heat calorimeter x (Tfinal-Tinitial)] = q(rxn)
Usually these are expressed as kcal/mol.
q/0.1 = cal/mol. Convert to kcal/mol.
The 0.1 comes from M x L = mols = 0.400 x 0.250 = 0.1 mol.
To find the enthalpy of neutralization for the reaction, we need to calculate the heat absorbed or released during the reaction. This can be determined using the equation:
q = mcΔT
where:
q is the heat change in Joules (J)
m is the mass of the system in kilograms (kg)
c is the specific heat capacity of the system in J/kg·K
ΔT is the change in temperature in Kelvin (K)
Let's break down the calculation step by step:
Step 1: Calculate the mass of the solutions
The total mass of the solutions is the sum of the volumes multiplied by their respective densities.
Density of water (assumed solution) = 1 g/cm³ = 1000 kg/m³ = 1 kg/L
Mass of NaOH solution = volume * concentration * density
= 0.25 L * 0.400 mol/dm³ * 1000 g/L
= 100 g
Mass of HCl solution = volume * concentration * density
= 0.25 L * 0.400 mol/dm³ * 1000 g/L
= 100 g
Step 2: Calculate the total mass of the system
The total mass of the system includes the mass of the solutions and the mass of the calorie meter.
Total mass = mass of NaOH solution + mass of HCl solution + mass of calorie meter
= 100 g + 100 g + 500 g
= 700 g
Step 3: Calculate the heat absorbed or released (q) using the equation q = mcΔT
We'll consider the heat released during the reaction since it's an exothermic process.
q = (mNaOH + mHCl + mcalorie meter) * ctotal * ΔT
where ctotal = (cmass solution + ccalorie meter)
ctotal = (msolution * csolution) + (mcalorie meter * ccalorie meter)
Given:
Specific heat capacity of the solution = 4200 J/kg·K
Specific heat capacity of the calorie meter = 400 J/kg·K
Change in temperature (ΔT) = 19.55 °C - 17.05 °C = 2.5 K
ctotal = ((200 g * 4200 J/kg·K) + (500 g * 400 J/kg·K))
= (840000 J/kg + 200000 J/kg)
= 1040000 J/kg
q = (700 g * 1040000 J/kg) * 2.5 K
= 1820000000 J
Step 4: Convert the heat change (q) to kilojoules (kJ) and find the moles of HCl (or NaOH) reacted.
1 kJ = 1000 J
1 mol HCl (or NaOH) = 0.4 moldm³ * 0.25 L = 0.1 mol
q = 1820000000 J = 1820000 kJ
Step 5: Calculate the enthalpy of neutralization
The enthalpy of neutralization (ΔH) represents the heat change per mole of HCl (or NaOH) reacted. Assuming the reaction is 1:1, our ΔH will be:
ΔH = q / n
where n = moles of HCl (or NaOH) reacted
ΔH = 1820000 kJ / 0.1 mol
= 18200000 kJ/mol
Therefore, the enthalpy of neutralization for the reaction is 18200000 kJ/mol.