Question #903e4

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Question #903e4

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Answer 1 · 2017-06-07T12:26:07

The enthalpy of solution is B. -57.61 kJ/mol.

Explanation:

There are two heats involved in this problem.

$heat of solution of KOH + heat absorbed by water = 0$ $"heat of solution of KOH + heat absorbed by water" = 0$
$m m m m m q_{1} m m m m m l l + m m m m m q_{2} m m m m m l = 0$ $color(white)(mmmmm)q_1color(white)(mmmmmll) +color(white)(mmmmm) q_2color(white)(mmmmml) = 0$
$color(white)(mmmm)nΔHcolor(white)(mmmmmmmml) + mCΔT color(white)(mmmml)= 0$

Let's calculate the heats separately.

$n = 13.90 color(red)(cancel(color(black)("g KOH"))) × "1 mol KOH"/(56.11 color(red)(cancel(color(black)("g KOH")))) = "0.2477 mol KOH"$

$q_1 = nΔH = 0.2477ΔH color(white)(l)"mol"$

$m = "(13.90 + 122.40) g" = "136.30 g"$
$C = "4.184 J·g·°C"^"-1"$
$ΔT = "(45.73 - 20.70) °C" = "25.03 °C"$

$q_2 = mCΔT = 136.30 color(red)(cancel(color(black)("g"))) × "4.184 J"·color(red)(cancel(color(black)("g"^"-1"·"°C"^"-1"))) × 25.03 color(red)(cancel(color(black)("°C"))) = "14 274 J" = "14.274 kJ"$

$q_1 + q_2 = 0.2477ΔH color(white)(l)"mol" + "14.274 kJ" = 0$

$ΔH = "-14.274 kJ"/"0.2477 mol" = "-57.62 kJ/mol"$

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Question #903e4

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