# How do I calculate the ΔH this reaction?

Aug 21, 2017

Δ_text(neut)H = "-55.8 kJ/mol".

#### Explanation:

There are two heat transfers involved in this problem.

$\text{heat of neutralization + heat gained by solution} = 0$
$\textcolor{w h i t e}{m m m m m} {q}_{1} \textcolor{w h i t e}{m m m m l l} + \textcolor{w h i t e}{m m m m m m} {q}_{2} \textcolor{w h i t e}{m m m m l} = 0$
color(white)(mmml)nΔ_text(neut)Hcolor(white)(mmm) +color(white)(mmmmll) mCΔTcolor(white)(mmm)= 0

(a) Heat of neutralization

The net ionic equation for the reaction is

$\text{H"_3"O"^"+" + "OH"^"-" → 2"H"_2"O}$

$\text{Moles of H"_3"O"^"+" = "0.020 60" color(red)(cancel(color(black)("L H"_2"SO"_4))) × (0.5000 color(red)(cancel(color(black)("mol H"_2"SO"_4))))/(1 color(red)(cancel(color(black)("L H"_2"SO"_4)))) × ("2 mol H"_3"O"^"+")/(1 color(red)(cancel(color(black)("mol H"_2"SO"_4)))) ="0.020 60 mol H"_3"O"^"+}$

$\text{Moles of OH"^"-" = "0.020 60" color(red)(cancel(color(black)("L NaOH"))) × (1.000 color(red)(cancel(color(black)("mol NaOH"))))/(1 color(red)(cancel(color(black)("L NaOH")))) × ("1 mol OH"^"-")/(1 color(red)(cancel(color(black)("mol NaOH")))) = "0.020 60 mol OH"^"-}$

You have equal moles of $\text{H"_3"O"^"+}$ and of $\text{OH"^"-}$, so you have $\text{0.020 60 mol reaction}$.

$n = \text{0.020 60 mol}$

q_1 = nΔ_text(neut)H = "0.020 60"Δ_text(neut)Hcolor(white)(l) "mol"

(b) Heat gained by water

$m = \text{(20.60 + 20.60) g" = "41.20 g}$
$C = \text{4.184 J·g·°C"^"-1}$
ΔT = T_2 - T_1 = "(30.17 - 23.50) °C" = "6.67 °C"

q_2 = mCΔT = 41.20 color(red)(cancel(color(black)("g"))) × "4.184J"·color(red)(cancel(color(black)("g"^"-1"·"°C"^"-1"))) × 6.67 color(red)(cancel(color(black)("°C"))) = "1150 J"

(c) Molar heat of neutralization

nΔ_text(neut)H + mCΔT = "0.020 60"Δ_text(neut)Hcolor(white)(l) "mol" + "1150 J" = 0

Δ_text(neut)H = "-1150 J"/"0.020 60 mol" = "-55 800 J/mol" = "-55.8 kJ/mol"