Question

When 2.714 g of AX (s) dissolves in 127.4 g of water in a coffee-cup calorimeter the temperature rises from 23.3 °C to 37.1 °C. Calculate the enthalpy change (in kJ/mol) for the solution process. AX(s) → A+(aq) + X-(aq) Assumptions for this calculation: The specific heat of the solution is the same as that of pure water (4.18 J/gK) The density of water = 1.000 g/mL The liquid’s final volume is not changed by adding the solid The calorimeter loses only a negligible quantity of heat. The formula weight of AX = 59.1097 g/mol. Be sure you include the correct sign for the enthalpy change.

Answer 1

Answer : The enthalpy change for the solution is 166.34 kJ/mol

Explanation :

First we have to calculate the enthalpy change of the reaction.

Formula used :

$\Delta H=mC\Delta T\\\\\Delta H=mC(T_2-T_1)$

where,

$\Delta H$ = change in enthalpy = ?

C = heat capacity of water = $4.18J/g.K$

m = total mass of sample = 2.174 + 127.4 = 129.6 g

$T_1$ = initial temperature = $23^oC=273+23=296K$

$T_2$ = final temperature = $37.1^oC=273+37.1=310.1K$

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

$\Delta H=mC(T_2-T_1)$

$\Delta H=129.6g\times 4.18J/g.K\times (310.1-296)K=7638.36J$

Now we have to calculate the moles of AX added to water.

$\text{ Moles of }AX=\frac{\text{ Mass of }AX}{\text{ Molar mass of }AX}=\frac{2.714g}{59.1097g/mole}=0.04592moles$

Now we have to calculate the enthalpy change for the solution.

As, 0.04592 moles releases heat = 7638.36 J

So, 1 moles releases heat = $\frac{7638.36}{0.04592}=166340.59J=166.34kJ$

Therefore, the enthalpy change for the solution is 166.34 kJ/mol