Answer:
ΔH of dissociation is 38,0 kJ/mol
Explanation:
The dissociation reaction of KBrO₃ is:
<em>KBrO₃ → K⁺ + BrO₃⁻ </em>
This dissolution consume heat that is evidenced with the decrease in water temperature.
The heat consumed is:
q = CΔTm
Where C is specific heat of water (4,186 J/mol°C)
ΔT is the temperature changing (18,0°C - 13,0°C = 5,0°C)
And m is mass of water (150,0 mL ≈ 150,0 g)
Replacing, heat consumed is:
q = 3139,5 J ≡ 3,14 kJ
13,8 g of KBrO₃ are:
13,8 g×(1mol/167g) = 0,0826 moles
Thus, ΔH of dissociation is:
3,14kJ / 0,0826mol = <em>38,0 kJ/mol</em>
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I hope it helps!
