When a 28.8mL of .500M H2SO4 is added to 28.8mL of 1.00M KOH in a coffee-cup calorimeter at 23.50 degrees Celsius, the temperature rises to 30.17 degrees Celsius. Calculate the ^H(change in enthalpy)of this reaction. (assume total volume is the sum of individual volumes and that the density and specific heat capacity of the solution are the same as pure water)

See my response to a similar problem earlier.

http://www.jiskha.com/display.cgi?id=1223794758

To calculate the change in enthalpy (ΔH) of this reaction, we can use the equation:

ΔH = q / moles

where q is the heat exchanged and moles is the number of moles of the limiting reactant used.

First, we need to determine the limiting reactant. From the given information, we know the initial volume of both H2SO4 and KOH is 28.8 mL. Since the molarity and volume are the same for both solutions, we can assume that they will react in a 1:1 ratio.

Next, we need to calculate the moles of the limiting reactant used.
Moles of H2SO4 = Molarity x Volume = 0.500 mol/L x 0.0288 L = 0.0144 mol
Moles of KOH = Molarity x Volume = 1.00 mol/L x 0.0288 L = 0.0288 mol

Since the reactants are in a 1:1 ratio, H2SO4 is the limiting reactant because it has fewer moles.

Now, we need to calculate the heat exchanged (q). We can do this using the equation:

q = m x C x ΔT

where m is the mass of the solution, C is the specific heat capacity of water, and ΔT is the change in temperature.

Since the density and specific heat capacity of the solution are the same as pure water, we can assume the solution has the same density and specific heat capacity as water. The density of water is approximately 1 g/mL. Therefore, the mass of the solution is:

Mass of solution = Volume of solution x Density of water = (28.8 mL + 28.8 mL) x 1 g/mL = 57.6 g

The specific heat capacity of water is 4.18 J/g°C. Therefore, the heat exchanged is:

q = 57.6 g x 4.18 J/g°C x (30.17°C - 23.50°C) = 1628.56 J

Finally, we can calculate the change in enthalpy:

ΔH = q / moles = 1628.56 J / 0.0144 mol = 113,028.33 J/mol (rounded to four significant figures)

Therefore, the change in enthalpy of this reaction is approximately 113,028.33 J/mol.