Calculate the molar heat of reaction for the NaOH(aq) using the data obtained from this experiment. (this is a neutralization question)

H2so4
Volume= 30ml
Concentration=1.0mol/l
Initial temp. = 29.9

NAOH
Volume= 50ml
C = 1.0 mol/L
Initial temp. = 29

Final temp of the two contents = 32.9

This is my solution:

El = Eg
NaOH H2O

nH = mc∆t

0.05mol x H = 0.08L x 4.19 x 3.9

H = 26.15

The standard molar heat of neutralization of sodium hydroxide,
is -57 kJ/mol.

so shouldn't my calculated value be close to the standard value? Please tell me whether my solution is right or wrong!

Thank you in advance =)

To calculate the molar heat of reaction for the neutralization between NaOH(aq) and H2SO4(aq), you are on the right track. However, there seems to be a mistake in your calculations.

Firstly, let's write the balanced equation for the neutralization reaction:

2NaOH(aq) + H2SO4(aq) -> Na2SO4(aq) + 2H2O(l)

Now, let's determine the heat exchanged during the reaction using the formula:

q = m * c * ∆T

where:
q = heat exchanged
m = mass (in grams) or moles of the substance
c = specific heat capacity (in J/(g·°C) or J/(mol·°C))
∆T = change in temperature (in °C)

For H2SO4:
* Volume = 30 mL = 0.03 L
* Concentration = 1.0 mol/L
* ∆T = (32.9°C - 29.9°C) = 3°C

nH2SO4 = Volume * Concentration = 0.03 L * 1.0 mol/L = 0.03 mol
mH2SO4 (in grams) = nH2SO4 * Molar Mass of H2SO4

For NaOH:
* Volume = 50 mL = 0.05 L
* Concentration = 1.0 mol/L
* ∆T = (32.9°C - 29°C) = 3.9°C

nNaOH = Volume * Concentration = 0.05 L * 1.0 mol/L = 0.05 mol
mNaOH (in grams) = nNaOH * Molar Mass of NaOH

Now, let's calculate the heat exchanged for H2SO4:
qH2SO4 = mH2SO4 * cH2SO4 * ∆T

Calculate the heat exchanged for NaOH:
qNaOH = mNaOH * cNaOH * ∆T

Finally, to determine the molar heat of reaction, we need to find the heat exchanged per mole of NaOH reacted. Since the balanced equation shows that the reaction of 2 moles of NaOH releases a certain amount of heat, we need to divide the calculated heat by 2:

∆Hrxn = (qH2SO4 + qNaOH) / (2 * nNaOH)

Compare this calculated value with the standard molar heat of neutralization (-57 kJ/mol) to check if they are close.