the enthalpy change when a strong acid is neutralized by base is -56.1 kJ/mol.

If 135 ml of .450 M HI at 23.15 C is mixed with 145 of .500 M NaOH at 23.15 C What will the max temperature reached by the resulting solution?
tamana: Assume the specific heat is 4.18 J/gC and dencity is the same as H2O

-10.4

To determine the maximum temperature reached by the resulting solution, we can use the equation:

q = mcΔT

where:
q is the heat transferred
m is the mass of the solution
c is the specific heat capacity of the solution
ΔT is the change in temperature

First, we need to calculate the mass of the solution. Assuming the density is the same as water, we can use the formula:

Density = Mass / Volume

Since we have the volumes of the acid and base, we can find the total volume and then use the density of water (which is approximately 1 g/mL) to calculate the mass.

Volume of HI (acid) = 135 mL
Volume of NaOH (base) = 145 mL

Total volume = Volume of the acid + Volume of the base
Total volume = 135 mL + 145 mL = 280 mL

Now, let's convert the total volume to grams:

Mass = Volume x Density
Mass = 280 mL x 1 g/mL
Mass = 280 g

Next, we need to calculate the moles of HI and NaOH. We can use the following equation:

moles = concentration x volume

Moles of HI = 0.450 M x 0.135 L
Moles of HI = 0.06075 moles

Moles of NaOH = 0.500 M x 0.145 L
Moles of NaOH = 0.0725 moles

Since HI and NaOH react in a 1:1 ratio, they will completely neutralize each other. Therefore, the limiting reagent is the reactant with the smaller number of moles, which in this case is HI.

The enthalpy change (ΔH) for the neutralization reaction is given as -56.1 kJ/mol. This means that 1 mole of HI reacts to release 56.1 kJ of energy.

Now, let's calculate the heat released by the neutralization reaction:

q = ΔH x moles of HI
q = -56.1 kJ/mol x 0.06075 mol
q = -3.407 kJ

To convert this value to joules, we need to multiply by 1000:

q = -3.407 kJ x 1000 J/kJ
q = -3407 J

Now, we can plug the values into the heat equation:

-3407 J = (280 g) x (4.18 J/g°C) x ΔT

Solving for ΔT:

ΔT = -3407 J / (280 g x 4.18 J/g°C)
ΔT ≈ -3.22 °C

Since the change in temperature is negative, it means the resulting solution will cool down by approximately 3.22 °C compared to the initial temperatures of the acid and base.

Therefore, the maximum temperature reached by the resulting solution would be approximately 23.15 °C - 3.22 °C = 19.93 °C.

To find the maximum temperature reached by the resulting solution, we can use the equation

q = mcΔT

where:
q is the heat transferred,
m is the mass of the solution,
c is the specific heat capacity of the solution, and
ΔT is the change in temperature.

To apply this equation, we need to calculate the mass of the solution. We have the volumes and concentrations of both HI and NaOH solutions, so we can use these values to find the mass.

First, let's calculate the mass of HI:
mass_HI = volume_HI * density_HI

Since the density of HI is not given but is stated to be the same as water, we can assume it to be 1 g/mL.

mass_HI = 135 ml * 1 g/mL
mass_HI = 135 g

Next, let's calculate the mass of NaOH:
mass_NaOH = volume_NaOH * density_NaOH

Assuming the density of NaOH is the same as water, which is 1 g/mL:

mass_NaOH = 145 ml * 1 g/mL
mass_NaOH = 145 g

Now, we can calculate the total mass of the resulting solution:
mass_total = mass_HI + mass_NaOH

mass_total = 135 g + 145 g
mass_total = 280 g

Now that we have the mass of the solution, we can use the equation q = mcΔT to find the heat transferred (q). Since the enthalpy change of neutralization is given (-56.1 kJ/mol), we can calculate the heat produced by the reaction between HI and NaOH.

First, we need to calculate the moles of HI and NaOH reacting. We can use the given concentrations and the volumes to find the moles of each compound.

moles_HI = volume_HI * concentration_HI

moles_HI = 135 ml * 0.450 mol/L
moles_HI = 60.75 mmol (millimoles)

moles_NaOH = volume_NaOH * concentration_NaOH

moles_NaOH = 145 ml * 0.500 mol/L
moles_NaOH = 72.5 mmol (millimoles)

Since the stoichiometry of the reaction between HI and NaOH is 1:1, the number of moles reacting is equal.

moles_react = moles_HI = moles_NaOH
moles_react = 60.75 mmol

Now, we can calculate the heat transferred (q) using the equation:

q = moles_react * enthalpy_change

Note: The enthalpy change is given in kJ/mol, so we need to convert the moles to mol before using it in the equation.

moles_react = 60.75 mmol = 0.06075 mol

q = 0.06075 mol * (-56.1 kJ/mol)
q = -3.42 kJ

Now, let's convert the heat transferred to joules:

q = -3.42 kJ * 1000 J/1 kJ
q = -3420 J

Finally, we can calculate the change in temperature (ΔT) using:

q = mcΔT

Since the specific heat capacity (c) is given as 4.18 J/g°C and the mass (m) of the solution is 280 g:

-3420 J = 280 g * 4.18 J/g°C * ΔT

ΔT = -3420 J / (280 g * 4.18 J/g°C)
ΔT ≈ -3.17 °C

The negative sign indicates a decrease in temperature. To find the maximum temperature reached by the resulting solution, we need to subtract this change in temperature from the initial temperature.

Max temperature = Initial temperature - ΔT
Max temperature = 23.15°C - (-3.17°C)
Max temperature ≈ 26.32°C

So, the maximum temperature reached by the resulting solution is approximately 26.32°C.

HI + NaOH ==> NaI + H2O

mmoles HI = 135 mL x 0.450 = 60.75
mmoles NaOH = 145 x 0.500 = 72.5
Obviously, the NaOH is is excess, therefore there will be formed 60.75 mmoles of the salt (said another way, the neutralization reaction occurred with 0.06075 moles of the acid/base.
heat produced = 56,100 J/mol x 0.06075 = 3408 J
Then plug that into
3408 = mass water x specific heat water x (Tfinal-Tinitial) and solve for Tfinal.
If you want to do a little closer job, you can note that you will produce an extra 0.06075 moles H2O which is about 1.09 grams and that can be added to the volume from the acid and base. Check my work.