The addition of 50mL of 2.02 M NaOH to 50 mL of 2.00 M HNO3 at 22.0 degree celsius resulted in maximum temperature of 36.1 degree celsius with a total mass of the solution of 100.0 g. calculate molar heat of neutralization and compare it to the results calculated from hess's law.
I know you have to use the equation delta h=-mc delta t
To calculate the molar heat of neutralization, we can use the equation q = mcΔT, where q is the heat released or absorbed, m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature.
Step 1: Calculate the heat released or absorbed during the reaction.
q = mcΔT
Given:
m = 100.0 g
ΔT = (36.1 - 22.0) °C = 14.1 °C
Step 2: Calculate the heat capacity of the solution.
The specific heat capacity of water is 4.18 J/g°C.
c = 4.18 J/g°C
Step 3: Substitute the values into the equation.
q = (100.0 g) (4.18 J/g°C) (14.1 °C)
q = 5877.840 J
Step 4: Convert the heat to kilojoules.
1 kJ = 1000 J
5877.840 J = 5.877840 kJ
Step 5: Calculate the moles of the limiting reagent.
The balanced equation for the neutralization reaction is:
HNO3 + NaOH -> NaNO3 + H2O
One mole of HNO3 reacts with one mole of NaOH, so the number of moles of HNO3 and NaOH is equal.
Given:
Volume of NaOH = 50 mL = 0.050 L
Concentration of NaOH = 2.02 M
moles of NaOH = (0.050 L) (2.02 mol/L) = 0.101 mol
Step 6: Calculate the molar heat of neutralization.
molar heat of neutralization = q / moles of NaOH
molar heat of neutralization = (5.877840 kJ) / (0.101 mol)
molar heat of neutralization = 58.183 kJ/mol
Now, we need to compare this result to the results calculated from Hess's law.
Hess's law states that the enthalpy change of a reaction is independent of the route taken. It means that the enthalpy change for a reaction can be calculated by taking the difference between the enthalpies of the reactants and products, regardless of the intermediate steps.
In this case, we need to compare the molar heat of neutralization obtained from the experimental data (58.183 kJ/mol) to the theoretical calculation using Hess's law.
To calculate the enthalpy change using Hess's law, we need the enthalpies of formation for the reactants and products. Without that information, we cannot directly compare the two values.
So, we can conclude that the molar heat of neutralization obtained experimentally is 58.183 kJ/mol, but we cannot compare it to the results calculated from Hess's law without additional information about the enthalpies of formation.
To calculate the molar heat of neutralization, we need to use the equation:
q = m×C×ΔT
where:
q = heat (in Joules)
m = mass of the solution (in grams)
C = specific heat capacity of the solution (assuming it is 4.18 J/g°C)
ΔT = change in temperature (final temperature - initial temperature)
First, let's calculate the heat released (q) in the reaction.
Step 1: Calculate the moles of NaOH:
Using the equation: Molarity (M) = moles (mol) / volume (L)
Given:
Volume of NaOH = 50 mL = 0.050 L
Molarity of NaOH = 2.02 M
moles of NaOH = Molarity × Volume
moles of NaOH = 2.02 M × 0.050 L
Step 2: Calculate the moles of HNO3:
Using the same equation as above:
Volume of HNO3 = 50 mL = 0.050 L
Molarity of HNO3 = 2.00 M
moles of HNO3 = Molarity × Volume
moles of HNO3 = 2.00 M × 0.050 L
Step 3: Determine the limiting reactant:
The reaction between NaOH and HNO3 is a 1:1 ratio, so the reactant with fewer moles is the limiting reactant.
In this case, both reactants have the same number of moles.
Since both reactants are in equimolar amounts, we can proceed with the calculation.
Step 4: Calculate the total moles of the solution:
Total moles of solution = moles of NaOH + moles of HNO3
Step 5: Calculate the mass of the solution:
Given: Total mass of the solution = 100.0 g
Step 6: Calculate the change in temperature (ΔT):
Given: Initial temperature = 22.0°C
Final temperature = 36.1°C
ΔT = Final temperature - Initial temperature
Now that we have gathered all the necessary information, we can calculate the heat released (q) using the equation mentioned at the beginning.
Once we have the value of q, we can use it to determine the molar heat of neutralization.
Comparing the experimental result to Hess's law can be done by calculating the theoretical value of the molar heat of neutralization using the enthalpy change of formation of the products and reactants of the neutralization reaction.