# Question 51bcc

Jul 22, 2017

The total amount of heat evolved is 5050 J; Δ_text(neut)H = "-37.4 kJ/mol".

#### Explanation:

There are three heat transfers involved in this problem.

$\text{heat of neutralization + heat gained by solution" + "heat gained by calorimeter} = 0$
${q}_{1} + {q}_{2} + {q}_{3} = 0$
nΔ_text(neut)H +mC_2ΔT + C_3ΔT= 0

Let's calculate the heats separately.

$n = \text{0.135 mol}$

q_1 = nΔ_text(neut)H = 0.135Δ_text(neut)Hcolor(white)(l) "mol"

$m = \text{(50.0 + 50.0) g" = "100.0 g}$
${C}_{2} = \text{4.184 J·g·°C"^"-1}$
ΔT = "(33.5 - 22.0) °C" = "11.5 °C"

q_2 = mC_2ΔT = 100.0 color(red)(cancel(color(black)("g"))) × "4.184J"·color(red)(cancel(color(black)("g"^"-1"·"°C"^"-1"))) × 11.5 color(red)(cancel(color(black)("°C"))) = "4812 J"

q_3 = C_3ΔT = 20.6 "J"·color(red)(cancel(color(black)("°C"))) × 11.5 color(red)(cancel(color(black)("°C"))) = "236.9 J"

${q}_{2} + {q}_{3}$ represents the heat from the neutralization that was absorbed by the water and the calorimeter.

${q}_{2} + {q}_{3} = \text{(4812 + 236.9) J = 5048 J}$

So, the total heat evolved is 5050 J (3 significant figures).

q_1 = 0.135Δ_text(neut)Hcolor(white)(l) "mol" = -(q_2 + q_3) = "-5048 J"

Δ_text(neut)H = "-5048 J"/"0.135 mol" = "-37 400 J/mol" = "-37.4 kJ/mol"#