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Makovka662 [10]
3 years ago
14

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?
Chemistry
1 answer:
mamaluj [8]3 years ago
8 0

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

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