When 25.0 mL of 1.0 M H2SO4

is added to 50.0 mL of 1.0 M NaOH at 25.0 °C in a calorimeter, the temperature of the aqueous solution
increases to 33.9 °C. Assuming the specific heat of the solution is 4.18 J/(g·°C), that its density is 1.00 g/mL, and that the calorimeter itself absorbs a negligible amount of heat, calculate H (in kJ) for the reaction:
H2SO4(aq) + 2 NaOH (aq) 2 H2O (l) + Na2SO4
(aq)

25ml+50ml=75ml *(1g/ml)= 75g

q= m*c*delta T
q= (75g) (4.18) (+8.9)
q= 2790.15 J = delta H
q= 2.79 KJ

25ml H2SO4
( x / 0.25ml) = 1 M H2SO4
x= 0.25 mol H2SO4
2.79KJ/ 0.25 mol H2SO4 = 11.16 KJ/mol H2SO4
delta H= - 11.2 kJ/mol
I'm not sure why it's negative but I hope this helps.

that took forever to figure out

change of heat of reaction= Vsolution/moles of limiting reactant X specific heat X change in temp X 1kj/1000j

It is negative because the temperature increases, therefore heat is liberated. This means it would be exothermic (delta H is -)

Did you know that chemistry jokes are sodium funny? But I'll try my best to help you with your question.

To calculate H for the reaction, we need to use the equation:
q = m * c * ΔT

Where:
q is the heat gained or lost by the solution,
m is the mass of the solution,
c is the specific heat capacity of the solution,
and ΔT is the change in temperature.

First, we need to find the mass of the solution. Given that the total volume of the solution is 75.0 mL and its density is 1.00 g/mL, the mass can be calculated as follows:
mass = volume * density
mass = 75.0 mL * 1.00 g/mL
mass = 75.0 g

Next, we need to calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 33.9 °C - 25.0 °C
ΔT = 8.9 °C

Now we can substitute the values into the equation:
q = 75.0 g * 4.18 J/(g·°C) * 8.9 °C

To convert this value to kJ, we divide by 1000:
q = 2.23 kJ

However, this value represents the heat gained or lost by the solution. Since the reaction is exothermic (releases heat), we can say that q = -H.

Therefore, H = -2.23 kJ.

Just remember, even though chemistry might be a complex subject, I'm here to make your experience a little more light-hearted!

To calculate the enthalpy change (ΔH) for the given reaction, we can use the equation:

ΔH = q / n

where:
- ΔH is the enthalpy change (in kJ)
- q is the heat absorbed or released (in J)
- n is the number of moles of the limiting reactant

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

q = m * c * ΔT

where:
- q is the heat (in J)
- m is the mass of the solution (in grams)
- c is the specific heat of the solution (in J/(g·°C))
- ΔT is the change in temperature (in °C)

1) Calculate the mass of the solution:
We have 25.0 mL of 1.0 M H2SO4 and 50.0 mL of 1.0 M NaOH. The density of the solution is given as 1.00 g/mL. Therefore, the mass of the solution can be calculated as follows:

Mass of H2SO4 solution = volume of H2SO4 solution * density
= 25.0 mL * 1.00 g/mL
= 25.0 g

Mass of NaOH solution = volume of NaOH solution * density
= 50.0 mL * 1.00 g/mL
= 50.0 g

Total mass of the solution = mass of H2SO4 solution + mass of NaOH solution
= 25.0 g + 50.0 g
= 75.0 g

2) Calculate the change in temperature (ΔT):
The temperature of the mixture increased from 25.0 °C to 33.9 °C. Therefore:

ΔT = Final temperature - Initial temperature
= 33.9 °C - 25.0 °C
= 8.9 °C

3) Calculate the heat (q):
Using the equation q = m * c * ΔT, we can substitute the given values:

q = 75.0 g * 4.18 J/(g·°C) * 8.9 °C
= 27961.5 J

4) Convert the heat (q) to kJ:
Since the answer is in kJ, we need to convert the heat value to kilojoules (kJ).

q = 27961.5 J / 1000
= 27.9615 kJ

Lastly, calculate the number of moles of the limiting reactant (n):
In this reaction, H2SO4 and NaOH react in a 1:2 ratio, which means that 2 moles of NaOH react with 1 mole of H2SO4. Since we have equal molar concentrations of H2SO4 and NaOH, the number of moles of H2SO4 and NaOH will be the same.

Using the volume and molar concentration of H2SO4:
Volume (L) = 25.0 mL / 1000 = 0.025 L
Moles of H2SO4 = Volume (L) * Molar Concentration (mol/L)
= 0.025 L * 1.0 mol/L
= 0.025 mol

Therefore, the number of moles of the limiting reactant (n) is 0.025 mol.

Now we can calculate the enthalpy change (ΔH) using the equation ΔH = q / n:

ΔH = 27.9615 kJ / 0.025 mol
= 1118.46 kJ/mol

Therefore, the enthalpy change (ΔH) for the reaction is 1118.46 kJ/mol.