When 20.0 mL of 0.535 M H2SO4 is added to 20.0 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.)
H per mole of H2SO4 reacted:
WebAssign will check your answer for the correct number of significant figures. Incorrect: Your answer is incorrect. . kJ/mol
H per mole of KOH reacted:
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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:
- ΔH is the enthalpy change per mole of reactant (in kJ/mol),
- q is the heat absorbed or released by the reaction (in J),
- n is the number of moles of the reactant that undergoes the reaction.
To find the heat absorbed or released by the reaction (q), we can use the formula:
q = m × c × ΔT
where:
- q is the heat absorbed or released by the reaction (in J),
- m is the mass of the solution (in g),
- c is the specific heat capacity of water (in J/g × K),
- ΔT is the change in temperature (in K).
First, let's calculate the mass of the solution:
The volume of H2SO4 added is 20.0 mL, and the density of water is 1.00 g/mL. Therefore, the mass of H2SO4 added is (20.0 mL) × (1.00 g/mL) = 20.0 g.
Similarly, the mass of KOH added is also 20.0 g.
Next, let's calculate the change in temperature (ΔT):
ΔT = final temperature - initial temperature
ΔT = 30.17°C - 23.50°C = 6.67°C
Now, let's calculate the heat absorbed or released by the reaction (q) for H2SO4 and KOH separately.
For H2SO4:
q(H2SO4) = m(H2SO4) × c × ΔT
q(H2SO4) = 20.0 g × 4.184 J/g×K × 6.67 K
For KOH:
q(KOH) = m(KOH) × c × ΔT
q(KOH) = 20.0 g × 4.184 J/g×K × 6.67 K
Next, let's calculate the number of moles of H2SO4 and KOH reacted:
For H2SO4:
n(H2SO4) = m(H2SO4) / M(H2SO4)
where M(H2SO4) is the molar mass of H2SO4.
Calculate the molar mass of H2SO4 (H=1.01 g/mol, S=32.07 g/mol, O=16.00 g/mol):
M(H2SO4) = (2 * 1.01 g/mol) + (1 * 32.07 g/mol) + (4 * 16.00 g/mol)
n(H2SO4) = 20.0 g / M(H2SO4)
For KOH:
n(KOH) = m(KOH) / M(KOH)
where M(KOH) is the molar mass of KOH.
Calculate the molar mass of KOH (K=39.10 g/mol, O=16.00 g/mol, H=1.01 g/mol):
M(KOH) = (1 * 39.10 g/mol) + (1 * 16.00 g/mol) + (1 * 1.01 g/mol)
n(KOH) = 20.0 g / M(KOH)
Finally, let's calculate the enthalpy change per mole of H2SO4 and KOH reacted:
ΔH(H2SO4) = q(H2SO4) / n(H2SO4)
ΔH(KOH) = q(KOH) / n(KOH)
Calculate the values for ΔH(H2SO4) and ΔH(KOH), and make sure to correctly convert the units to kJ/mol and round the answer to the correct number of significant figures as required.