In a constant-pressure calorimeter, 75.0 mL of 0.830 M H2SO4 was added to 75.0 mL of 0.310 M NaOH. The reaction caused the temperature of the solution to rise from 23.20 °C to 25.31 °C. If the solution has the same density and specific heat as water, what is ΔH for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes.
Oh boy, let's break this down!
First off, H2SO4 and NaOH are having a little party in the calorimeter. They're mixing together, releasing some heat, and causing the temperature to rise. How scandalous!
To find ΔH for this reaction, we need to remember that it's the heat exchanged at constant pressure. So we can use the equation:
q = m × C × ΔT
q represents the heat exchanged, m is the mass (which we can calculate from the densities and volumes of the solutions), C is the specific heat capacity, and ΔT is the change in temperature.
But wait, didn't they mention that the solution has the same density and specific heat as water? Well, that makes things a lot easier! We can use the densities to find the masses, and the specific heat capacity of water is 4.18 J/g°C.
So, let's calculate the masses first. The densities of the solutions are the same as water, which is approximately 1.00 g/mL. That means we have 75.0 g of H2SO4 and 75.0 g of NaOH. Taking into account the molar masses of H2SO4 (98.09 g/mol) and NaOH (40.00 g/mol), we can calculate the moles of each substance:
moles of H2SO4 = 75.0 g / 98.09 g/mol
moles of NaOH = 75.0 g / 40.00 g/mol
Now we need to figure out which of these two substances is the limiting reactant. The mole ratio between H2SO4 and NaOH is 1:2, so we can find out how many moles of H2SO4 react with NaOH by multiplying the moles of NaOH by the ratio:
moles of H2SO4 reacting = (moles of NaOH) × (1 mole H2SO4 / 2 moles NaOH)
Now we know the moles of H2SO4 that reacted, and that tells us the moles of water produced since the balanced chemical equation is 1:1 for H2SO4 : H2O.
Finally, we can use the equation for q with the calculated values to find ΔH:
ΔH = q / moles of water produced
After plugging in all those numbers and doing the math, you'll find the answer. But remember, this is a constant-pressure calorimeter, so ΔH represents the heat transfer per mole of water produced.
I hope that helps and puts a smile on your face!
To determine the enthalpy change (ΔH) for the reaction, we need to calculate the heat released or absorbed during the reaction. We can use the equation:
q = m * c * ΔT
where:
q = heat transferred (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)
First, we need to find the mass of the solution. Since the density and specific heat capacity of the solution are the same as water, we can assume that the density is 1 g/mL. Therefore, the mass of the solution is:
mass = volume * density
= (75.0 mL + 75.0 mL) * 1 g/mL
= 150.0 g
Next, we calculate the change in temperature:
ΔT = final temperature - initial temperature
= 25.31 °C - 23.20 °C
= 2.11 °C
Now, we can calculate the heat released or absorbed during the reaction:
q = m * c * ΔT
Assuming the specific heat capacity of water is 4.18 J/g°C:
q = 150.0 g * 4.18 J/g°C * 2.11 °C
= 1350.09 J
The heat released during the reaction is 1350.09 J.
To find the enthalpy change (ΔH) per mole of H2O produced, we need to convert the given concentrations of the reactants to moles.
moles of H2SO4 = volume of H2SO4 * concentration of H2SO4
= 75.0 mL * 0.830 mol/L
= 62.25 mmol
= 0.06225 mol
moles of NaOH = volume of NaOH * concentration of NaOH
= 75.0 mL * 0.310 mol/L
= 23.25 mmol
= 0.02325 mol
From the balanced chemical equation:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
We can see that 1 mole of H2SO4 produces 2 moles of H2O.
Therefore, moles of H2O produced = 2 * moles of H2SO4
= 2 * 0.06225 mol
= 0.1245 mol
Finally, we calculate the enthalpy change (ΔH) per mole of H2O produced:
ΔH = q / moles of H2O produced
= 1350.09 J / 0.1245 mol
= 10848 J/mol
The ΔH for the reaction (per mole of H2O produced) is 10848 J/mol.