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Stells [14]
3 years ago
11

When a 3.25 g sample of solid sodium hydroxide was dissolved in a calorimeter in 100.0 g of water, the temperature rose from 23.

9 °C to 32.0 °C. Calculate ∆H (in kJ/mol) for the solution process: NaOH (s) → Na+ (aq) + OH- (aq)
Use a calorimeter heat capacity of Ccal = 15.8 J/°C
Chemistry
1 answer:
sertanlavr [38]3 years ago
4 0

Answer : The enthalpy change for the solution is 42.8 kJ/mol

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

q=[q_1+q_2]

q=[c_1\times \Delta T+m\times c_2\times \Delta T]

where,

q = heat released by the reaction

q_1 = heat absorbed by the calorimeter

q_2 = heat absorbed by the water

c_1 = specific heat of calorimeter = 15.8J/^oC

c_2 = specific heat of water = 4.18J/g^oC

m = mass of water = 100.0 g

\Delta T = change in temperature = T_2-T_1=(32.0-23.9)=8.1^oC

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

q=[(15.8J/^oC\times 8.1^oC)+(100.0g\times 4.18J/g^oC\times 8.1^oC)]

q=3513.8J=3.5138kJ        (1 kJ = 1000 J)

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

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

where,

\Delta H = enthalpy change = ?

q = heat released = 3.5138 kJ

m = mass of NaOH = 3.25 g

Molar mass of NaOH = 40 g/mole

\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{3.25g}{40g/mole}=0.0812mole

Now,

\Delta H=\frac{3.5138kJ}{0.0821mole}=42.8kJ/mol

Therefore, the enthalpy change for the solution is 42.8 kJ/mol

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