Question #1c0c4

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Answer 1 · 2015-03-30T17:06:00

$ΔH_"rxn" = "-61 kJ/mol"$

There are three heats to consider:

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

$q_1 + q_2 + q_3 = 0$

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

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

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

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

HA + B → BH⁺ + A⁻

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

$q_1 = nΔH = 0.050ΔH "mol"$

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

$0.050ΔH "mol" = "-3067 J"$

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