When 33.5 mL of 0.535 M H2SO4 is added to 33.5 mL of 1.07 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate H of this reaction per mole of H2SO4 and KOH reacted. (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 = 1.00 g/mL and c = 4.184 J/g×K.)

Question

When 33.5 mL of 0.535 M H2SO4 is added to 33.5 mL of 1.07 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate H of this reaction per mole of H2SO4 and KOH reacted. (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 = 1.00 g/mL and c = 4.184 J/g×K.)

Answer 1

To calculate the enthalpy change (ΔH) of the reaction per mole of H2SO4 and KOH reacted, we can use the formula:

ΔH = q / n

where q is the heat transferred and n is the number of moles of the limiting reactant.

To find the number of moles of H2SO4 and KOH reacted, we can use the equation:

n = C × V

where C is the concentration of the solution and V is the volume in liters.

First, let's calculate the number of moles of H2SO4:

C(H2SO4) = 0.535 M
V(H2SO4) = 33.5 mL = 33.5 / 1000 L = 0.0335 L

n(H2SO4) = C(H2SO4) × V(H2SO4)
= 0.535 M × 0.0335 L
= 0.0179675 mol

Now, let's calculate the number of moles of KOH:

C(KOH) = 1.07 M
V(KOH) = 33.5 mL = 33.5 / 1000 L = 0.0335 L

n(KOH) = C(KOH) × V(KOH)
= 1.07 M × 0.0335 L
= 0.035845 mol

Since the reaction is a 1:1 ratio between H2SO4 and KOH, the number of moles of the limiting reactant is 0.0179675 mol. Therefore, we can use this value for the calculation of ΔH.

Next, we need to calculate the heat transferred (q). We can use the formula:

q = m × c × ΔT

where m is the mass of the solution and ΔT is the change in temperature.

First, let's calculate the mass of the solution:

density (d) = 1.00 g/mL
volume (V) = 33.5 mL + 33.5 mL = 67 mL = 67 g

m(solution) = d × V
= 1.00 g/mL × 67 mL
= 67 g

Now, let's calculate ΔT:

ΔT = T(final) - T(initial)
= 30.17°C - 23.50°C
= 6.67°C

Finally, we can calculate the heat transferred (q):

q = m × c × ΔT
= 67 g × 4.184 J/g×K × 6.67°C
= 1868.14 J

Now we can calculate the enthalpy change (ΔH) per mole of H2SO4 and KOH reacted:

ΔH = q / n
= 1868.14 J / 0.0179675 mol
= 1.04 x 10^5 J/mol

Therefore, the enthalpy change (ΔH) of this reaction per mole of H2SO4 and KOH reacted is 1.04 x 10^5 J/mol.