Respuesta :
Explanation:
The given data is as follows.
Molarity of [tex]H_{2}SO_{4}[/tex] = 0.500 M
Volume [tex]H_{2}SO_{4}[/tex] solution = 21.7 mL
Hence, calculate the number of moles of sulfuric acid as follows.
No. of moles of [tex]H_{2}SO_{4}[/tex] = [tex]molarity \times volume[/tex]
= [tex]0.500 M \times 21.7 \times 10^{-3} L[/tex]
= 0.01085 mol
Now, calculate the moles of KOH as follows.
No. of moles of KOH = [tex]\text{molarity of KOH} \times \text{volume KOH in liter}[/tex]
= [tex]1.00 M \times 21.7 \times 10^{-3} L[/tex]
= 0.0217 mol
As the reaction equation is as follows.
[tex]H_{2}SO_{4}(aq) + 2KOH(aq) \rightarrow K_{2}SO_{4}(aq) + 2H_{2}O (l)[/tex]
Calculate the number of moles of water as follows.
No. of moles of [tex]H_{2}O[/tex] formed = [tex]2 \times moles of H_{2}SO_{4}[/tex] = (moles of KOH)
= 0.0217 mol
Now, total volume of the solution is as follows.
Total volume of solution = (volume [tex]H_{2}SO_{4}[/tex]) + (volume KOH)
= (21.7 mL) + (21.7 mL)
= 43.4 mL
Total mass of solution = [tex](\text{Total volume of solution}) \times (\text{density of solution})[/tex]
Total mass of solution = [tex]43.4 mL \times 1.00 g/mL[/tex]
= 43.4 g
Heat change of solution = [tex](\text{Total mass of solution}) \times (\text{specific heat of solution}) \times (\text{final temperature - initial temperature})[/tex]
= [tex](43.4 g) \times (4.184 J/g^{o}C) \times (30.17^{o}C - 23.50^{o}C)[/tex]
= 1211.18 J
As, heat change of reaction = -(Heat change of solution)
Therefore, heat change of reaction = -1211.18 J
Hence, calculate the change in enthalpy as follows.
[tex]\DeltaH = \frac{\text{Heat change of reaction}}{\text{moles of H_{2}O formed}}[/tex]
= [tex]\frac{-1211.18 J}{0.0217 mol}[/tex]
= -55814.56 J/mol
or, = -55.8 kJ/mol (as 1 kJ = 1000 J)
Thus, we can conclude that [tex]\Delta H[/tex] of the given reaction is -55.8 kJ/mol.
