Respuesta :
Answer:
The ΔH of this reaction is 55.8 kJ/mol.
Explanation:
[tex]Molarity=\frac{Moles}{Volume}[/tex]
Moles of sulfuric acid:
Moles of sulfuric acid = [tex]0.500 mol/L\times 0.026 L=0.013 mol[/tex]
Moles of Potassium hydroxide:
Moles of Potassium hydroxide= [tex]1.00 mol/L\times 0.026 L=0.026 mol[/tex]
[tex]H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O,\Delta H_r=?[/tex]
1 mol of sulfuric acid reacts witrh 2 mol of potassium hydroxide.
Then 0.013 mol of sulfuric acid will react with:
[tex]0.013\times \frac{2}{1}=0.026 mol[/tex] of potassium hydroxide.
Total volume of the solution = 26.0 mL+26.0 mL= 52 .0 mL
The density of the solution is same as pure water = 1.00g/mL (given)
Mass of the solution ,m=[tex]Density\times Volume=1.00 g/mL\times 52.0 ml=52.000 g[/tex]
The specific heat capacity of the solution is same as pure water:
c =4.184 J/g°C(given)
Change in temperature of the solution = ΔT =30.17°C - 23.50°C=6.67°C
Heat released during the mixing of both the solution; Q
[tex]Q=mc\Delta T=52.000 g\times 4.184 J/g^oC\times 6.67^oC[/tex]
Q =1,451.17 J = 1.4511 kJ
When 0.013 mol of sulfuric acid reacts with 0.026 moles of potassium to give 0.026 moles of water.
1.4511 kJ of heat is released when 0.026 moles of water are formed.
Then , for 1 mole of water the energy release will be:
[tex]\frac{1.4511 kJ}{0.026 mol}=55.8 kJ[/tex]
So, the ΔH of this reaction is 55.8 kJ/mol.
The enthalpy of reaction is -55.8 KJ/mol.
From the equation of the reaction;
2KOH(aq) + H2SO4(aq) ------> K2SO4(aq) + 2H2O(l)
Number of moles of H2SO4 = 26.0/1000 × 0.500 M = 0.013 moles
Number of moles of KOH = 26.0/1000 × 1.00 M = 0.026 moles
2 moles of KOH produces 2 moles of water
Hence 0.0026 moles of KOH produces 0.026 moles of water.
Total volume of solution = 26.0 mL + 26.0 mL = 52 mL
Mass of water = density × volume = 1.00 g/mL × 52 mL = 52 g
Using the formula;
ΔH = mcθ
Mass of solution (m) = 52 g
Specific heat capacity of solution (c) = 4.184 J/g·°C
Temperature difference(θ) = 30.17°C - 23.50°C = 6.67°C
Substituting values;
ΔH = -() 52 g × 4.184 J/g·°C × 6.67°C/ 0.026 moles
ΔH = -(1.45 KJ/0.026 moles)
ΔH = -55.8 KJ/mol
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