In a constant‑pressure calorimeter, 75.0mL of 0.950 M H2SO4 was added to 75.0 mL of 0.250 M NaOH. The reaction caused the temperature of the solution to rise from 23.24∘C to 24.94∘C. If the solution has the same density and specific heat as water (1.00 g/mL and 4.184 J/(g⋅°C), respectively), what is ΔH for this reaction (per mole of H2O produced)?
Did you make a typo with H2SO4 and M of 0.950? I have worked the problem as if this were not a typo. If you meant it to be 0.250 all the following still applies.
H2SO4 + 2NaOH ==> 2H2O + Na2SO424.94 - 23.
mols NaOH = L x M = 0.075 x 0.250= 0.01875
mols H2O produced = 0.01875 = 0.01875
Total volume is 75 mL + 75 mL = 150 mL= 150 grams.
q = heat released = mass H2O x specific heat H2O x (Tfinal-Tinitial) =
150 x 4.184 x (24.94 - 23.24) = I estimate 1000 J but you need to recalculate this and all math in the problem. I've just estimated.
That is approximately 1000 J for 0.0375 mols H2O produced.
dH = about 1000/0.01875 = about 55,000 J or about 55 kJ/mol H2O.
Post your work if you get stuck.
I didn't mean to overwrite the post by Bob P but the results of each will give you the same answer.
To calculate the value of ΔH (the enthalpy change) for this reaction, we can use the equation:
ΔH = q / n
Where ΔH is the enthalpy change, q is the heat absorbed or released during the reaction, and n is the number of moles involved in the reaction.
To calculate q, we can use the equation:
q = m * c * ΔT
Where q is the heat absorbed or released, m is the mass of the solution, c is the specific heat capacity, and ΔT is the change in temperature.
First, let's calculate the mass of the solution:
mass = volume * density
The volumes of both H2SO4 and NaOH are given as 75.0 mL each, and the density of water is 1.00 g/mL. Therefore, the mass of the solution is:
mass = (75.0 mL H2SO4 + 75.0 mL NaOH) * 1.00 g/mL
Next, let's calculate the heat absorbed or released (q):
q = m * c * ΔT
m = mass of the solution (calculated above)
c = specific heat capacity of water (4.184 J/(g⋅°C))
ΔT = change in temperature (24.94 - 23.24)
Now, let's calculate the number of moles involved in the reaction:
n = mole ratio of water to the limiting reactant
In this case, the balanced equation for the reaction between H2SO4 and NaOH is:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
From the balanced equation, we can see that 2 moles of water are produced for every 1 mole of H2SO4. So, the mole ratio is 2:1.
Now, calculate the number of moles (n):
n = 75.0 mL H2SO4 * 0.950 mol/L * (1 L / 1000 mL)
Finally, substitute the values into the equation ΔH = q / n to find the enthalpy change per mole of H2O produced.
Moles H2SO4= .075*.950=.87
moles NaOH= .075*.250=.0187
reaction 2NaOH+H2SO4>>>2H2O+ Na2SO4
so you need two moles NaOH for every mole of sulfuric, so the reaction has excess acid, and you formed .0187 moles water, the product.
heat lost by reaction- massfluid*c*changetemp=.150*4.18*1.70= you do it, check units.
Hr= heat lost /moles product