Question

2. When 15.3 g of sodium nitrate, NaNO3,was dissolved in water in a calorimeter, the temperature fell from 25.00oC to 21.56oC. If the heat capacity of the solution plus the calorimeter is 1071 J/o C, what is the enthalpy change when one mole of NaNO3 dissolves

Answer 1

Answer: The enthalpy of the reaction is 20.46 kJ/mol

Explanation:

To calculate the heat absorbed by the calorimeter, we use the equation:

$q=c\Delta T$

where,

q = heat absorbed

c = heat capacity of calorimeter = 1071 J/°C

$\Delta T$ = change in temperature = $T_2-T_1=(21.56-25.00)^oC=-3.44^oC$

Putting values in above equation, we get:

$q=1071J/^oC\times (-3.44)^oC=-3684.24J=-3.684kJ$

Heat released by the calorimeter will be equal to the heat absorbed by the reaction.

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of sodium nitrate = 15.3 g

Molar mass of sodium nitrate = 85 g/mol

Putting values in above equation, we get:

$\text{Moles of sodium nitrate}=\frac{15.3g}{85g/mol}=0.18mol$

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat absorbed = 3.684 kJ

n = number of moles of sodium nitrate = 0.18 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{3.684kJ}{0.18mol}=20.46kJ/mol$

Hence, the enthalpy of the reaction is 20.46 kJ/mol

Answer 2

The enthalpy change when one mole of NaNO3 dissolves is 114  kJ/mol.

Molar mass of of sodium nitrate = 85 g/mol

The number of moles of sodium nitrate (n) = mass/ molar mass = 15.3 g/85 g/mol

= 0.18 moles

Temperature drop(dT) = 21.56oC - 25.00oC = -3.44oC

Heat capacity of solution plus the calorimeter = 1071 J/o C

We know that heat absorbed = cdT

c = heat capacity of solution plus the calorimeter

dT = temperature drop =  -3.44oC

H = 1071 J/o C ×  -3.44oC = -3.68 kJ

Now;

Heat of reaction = -( -3.68 kJ/ 0.18 moles) = 20.44 kJ/mol

If 0.18 moles has an enthalpy change of 20.44 kJ/mol

1 mole would have an enthalpy change of 20.44 kJ/mol × 1 mole/ 0.18 moles

= 114  kJ/mol

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