Chemistry, 23.10.2019 17:00 mrgandollins5222

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.

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Answer from: hopeschr2019

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

Answer from: reticentrobbie

When 2.714 g of AX(s) dissolves in 127.4 g of water in a coffee-cup calorimeter and the temperature rises from 23.3 °C to 37.1 °C, the enthalpy change for the solution process is -163 kJ/mol.

A coffee cup calorimeter is a constant-pressure calorimeter. As such, the heat that is measured in such a device is equivalent to the change in enthalpy.

A solution is prepared by dissolving 2.714 g of AX (solute) in 127.4 g of water (solvent). The mass of the solution (m) is:

m = 2.714g + 127.4 g = 130.1 g

When AX is dissolved, the temperature of the system rises from 23.3 °C to 37.1 °C. Assuming the solution has the same specific heat as water (c = 4.18 J/g.K = 4.18 J/g.°C), we can calculate the heat absorbed by the calorimeter using the following expression.

$Q = c \times m \times \Delta T = \frac{4.18J}{g. \° C } \times 130.1 g \times (37.1 \° C - 23.3 \° C) = 7.50 \times 10^{3} J = 7.50 kJ$

According to the law of conservation of energy, the sum of the heat absorbed by the calorimeter (Qcal) and the heat released by the solution (Qsol) is zero.

$Qcal + Qsol = 0\\Qsol = -Qcal = -7.50 kJ$

2.714 g of AX(s) with a molar mass of 59.1097 g/mol are dissolved. The corresponding moles are:

$2.714 g \times \frac{1mol}{59.1097g} = 0.04591 mol$

Finally, we can calculate the enthalpy change (in kJ/mol) for the solution process using the following expression.

$\Delta Hsol = \frac{Qsol}{n} = \frac{-7.50 kJ}{0.04591mol} = -163 kJ/mol$

When 2.714 g of AX(s) dissolves in 127.4 g of water in a coffee-cup calorimeter and the temperature rises from 23.3 °C to 37.1 °C, the enthalpy change for the solution process is -163 kJ/mol.

You can learn more about calorimetry here:

When 2.714 g of ax (s) dissolves in 127.4 g of water in a coffee-cup calorimeter the temperature ris

Answer from: Quest

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Answer from: Quest

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