When 22.50 mL of 0.500 M H2SO4 is added to 22.50 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate ΔH of this reaction. (Assume that the total volume is the sum of the individual volumes and that the density and specific heat capacity of the solution are the same as for pure water.) (d for water= 1.00 g/mL; c for water =4.184 J/g*C)

q rxn = mass H2O x specific heat H2O x (Tfinal-Tinitial)

mass H2O = 22.5 + 22.5 = ?
sp. h. is given
So q = approx 1200 (but that's just an estimate)and it is approx -1200 J since the reaction is exothermic.
That is delta H per 0.0225 mols.
Then dH in J/mol = approx -1200/0.0225 = about -53,000 J/mol. Convert to kJ/mol

To calculate the ΔH of this reaction, we need to use the equation:

ΔH = q / moles

First, let's calculate the moles of H2SO4 and KOH used in the reaction.

moles H2SO4 = volume H2SO4 x molarity H2SO4
moles H2SO4 = 22.50 mL x (0.500 mol/L / 1000 mL/mol)
moles H2SO4 = 0.01125 mol

moles KOH = volume KOH x molarity KOH
moles KOH = 22.50 mL x (1.00 mol/L / 1000 mL/mol)
moles KOH = 0.02250 mol

Next, let's calculate the heat (q) released or absorbed by the reaction using the formula:

q = mass x specific heat capacity x ΔT

The mass can be found using:

mass = volume x density
mass = (22.50 mL + 22.50 mL) x 1.00 g/mL
mass = 45 g

ΔT = final temperature - initial temperature
ΔT = 30.17°C - 23.50°C
ΔT = 6.67°C

Now, we can calculate q:

q = 45 g x 4.184 J/g·°C x 6.67°C
q = 1250.933 J

Finally, we can calculate ΔH:

ΔH = q / moles
ΔH = 1250.933 J / (0.01125 mol + 0.02250 mol)
ΔH = 1250.933 J / 0.03375 mol
ΔH = 37037.93 J/mol (rounded to four significant figures)

Therefore, the ΔH of this reaction is 37037.93 J/mol.

To calculate the enthalpy change (ΔH) of the reaction, we need to use the equation:

ΔH = q/m

where:
- ΔH is the enthalpy change of the reaction in units of joules (J)
- q is the heat transfer in units of joules (J)
- m is the mass in grams (g) of the solution

To find q, we can use the equation:

q = m × c × ΔT

where:
- m is the mass in grams (g) of the solution
- c is the specific heat capacity of water, which is given as 4.184 J/g°C
- ΔT is the change in temperature in degrees Celsius (°C)

Now, let's calculate the values for each part of the equation:

1. Calculate the mass (m) of the solution:
The total volume of the solution is the sum of the individual volumes of the H2SO4 and KOH. Since the density (d) of water is given as 1.00 g/mL, we can find the mass (m) using the equation:

m = V × d

For H2SO4:
m(H2SO4) = V(H2SO4) × d = 22.50 mL × 1.00 g/mL
m(H2SO4) = 22.50 g

For KOH:
m(KOH) = V(KOH) × d = 22.50 mL × 1.00 g/mL
m(KOH) = 22.50 g

The total mass (m) of the solution is the sum of the masses of H2SO4 and KOH:
m(total) = m(H2SO4) + m(KOH)
m(total) = 22.50 g + 22.50 g
m(total) = 45.00 g

2. Calculate the change in temperature (ΔT):
ΔT = T(final) - T(initial)
ΔT = 30.17°C - 23.50°C
ΔT = 6.67°C

3. Calculate the heat transfer (q):
q = m × c × ΔT
q = 45.00 g × 4.184 J/g°C × 6.67°C
q = 1248.438 J

Finally, we can substitute the values of q and m into the equation for ΔH:

ΔH = q/m
ΔH = 1248.438 J / 45.00 g
ΔH = 27.74 J/g

Therefore, the enthalpy change (ΔH) of this reaction is 27.74 J/g.

for the given volumes

(22.5 + 22.5) * (30.17 - 23.50) * 1.00
... * 4.184