Question

Question #1c0c4

Answer 1

`ΔH_"rxn" = "-61 kJ/mol"`

There are three heats to consider:

heat of reaction + heat to warm solution + heat to warm calorimeter = 0

`q_1 + q_2 + q_3 = 0`

`q_3 = C_"Cal"ΔT = "12.0 J"cancel("°C⁻¹") × 6.5 cancel("°C") = "78 J"`

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`V = "(50.0 +50.0) mL" = "100.0 mL"`

`m = 100.0 cancel("mL") × "1.10 g"/(1 cancel("mL")) = "110 g"`

`q_2 = mcΔT = 110 cancel("g") × "4.18 J"·cancel("g⁻¹°C⁻¹") × 6.5 cancel("°C") = "2989 J"`

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HA + B → BH⁺ + A⁻

`n_"HA" = n_"B" = 0.0500 cancel("L") × "1.0 mol"/(1 cancel("L")) = "0.050 mol"`

`q_1 = nΔH = 0.050ΔH "mol"`

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`q_1 + q_2 + q_3 = 0.050ΔH " mol + 2989 J + 78 J" = 0`

`0.050ΔH "mol" = "-3067 J"`

`ΔH = "-3067 J"/"0.050 mol" = "-61 000 J" = "-61 kJ"`