Question

An amount of solid barium chloride, 20.8 g, is dissolved in 100 g water in a coffee-cup calorimeter by the reaction: BaCl2 (s)  Ba2+(aq) + 2Cl−(aq) The water is originally at 25.0 °C and after the reaction the temperature of the solution is 26.6 °C. (Cs = 4.04 J/(g°C) for the solution). What is the enthalpy change (ΔH) associated with the reaction as written?

Answer 1

Answer : The enthalpy change during the reaction is -6.48 kJ/mole

Explanation :

First we have to calculate the heat gained by the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

m = mass of water = 100 g

c = specific heat = $4.04J/g^oC$

$T_{final}$ = final temperature = $26.6^oC$

$T_{initial}$ = initial temperature = $25.0^oC$

Now put all the given values in the above formula, we get:

$q=100g\times 4.04J/g^oC\times (26.6-25.0)^oC$

$q=646.4J$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles barium chloride = $\frac{\text{Mass of barium chloride}}{\text{Molar mass of barium chloride}}=\frac{20.8g}{208.23g/mol}=0.0998mole$

$\Delta H=-\frac{646.4J}{0.0998mole}=-6476.95J/mole=-6.48kJ/mole$

Therefore, the enthalpy change during the reaction is -6.48 kJ/mole

Answer 2

The reaction of Barium chloride in the calorimeter has the value of change in enthalpy of 6.47 kJ.

What is specific heat?

The specific heat cam be defined as the amount of energy required to raise the temperature of the substance by 1 degree Celsius.

The heat required ($q$) by the reaction can be given as:

$q=mc\;\Delta T$

Where, the mass of the barium chloride, $m=100\;\rm g$

The specific heat of the calorimeter solution, $c=4.04\;\rm J/g.^\circ C$

The change in temperature ($\Delta T$) of the calorimeter is given as:

$\Delta T=26.6-25\;^\circ \text C\\\Delta T=1.6\;^\circ \text C$

Substituting the values for the heat of the reaction:

$q=\rm 100\;g\;\times\;4.04\;J/g^\circ C\;\times\;1.6\;^\circ C\\\textit q=646.4\;J$

The heat of the reaction is 646.4 J.

The moles of barium chloride ($n$) utilized in the reaction are:

$\rm Moles\;(\textit n)=\dfrac{Mass}{Molar\;mass} \\\\\textit n=\dfrac{20.8\;g}{208.23\;g/mol}\\\\\textit n=0.0998\;mol$

The change in enthalpy ($\Delta H$) in the reaction of barium chloride is given as:

$\Delta H=\dfrac{q}{n}\\\\ \Delta H=\dfrac{646.4\;\text J}{0.0988\;\text{mol}} \\\\\Delta H=6476\;\text J\\\\\Delta H=6.47\;\rm kJ$

The change in enthalpy of the reaction of barium chloride is 6.47 kJ.

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