Chemistry, 29.10.2019 19:31 yasameenakbari

From the calorimetric data, calculate δh for the reaction that occurs on mixing. assume that the calorimeter absorbs only a negligible quantity of heat, that the total volume of the solution is 100.0 ml, and that the specific heat and density of the solution after mixing are the same as that of pure water.

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Answer from: jennyn26

Explanation:

It is assumed that concentration of given $CuSO_{4}$ is 1.0 M and its volume is 50.0 ml or $50 \times 10^{-3}$ L (as 1 L = 1000 mL).

For KOH, concentration is 2.00 M and volume is 50.0 ml or $50 \times 10^{-3}$ L.

$T_{1}$ = $21.5^{o}C$ and $T_{2}$ = $27.7^{o}C$

Therefore, the given reaction equation will be as follows.

$CuSO_{4}(aq) + 2KOH (aq) \rightarrow Cu(OH)_{2}(s) + K_{2}SO_{4} (aq)$

Hence, calculate the moles of copper sulfate as follows.

Moles of $CuSO_{4} = molarity \times volume$

= $1.00 M \times 50.0 \times 10^{-3} L$

= 0.05 mol

Also, the number of moles of KOH will be as follows.

Moles of KOH = (molarity) × (volume)

= $2.00 M \times 50.0 \times 10^{-3} L$

= 0.10 mol

Therefore, moles $Cu(OH)_{2}$ formed = moles $CuSO_{4}$ consumed

0.5 × (moles of KOH consumed)

$0.5 \times 0.10 mol$

= 0.05 mol

It is given that volume of solution = 100.0 mL

Hence, mass of solution = (Volume of solution) × (density of solution)

= (100.0 mL) × (1.00 g/mL)

= 100.0 g

Also, heat absorbed by solution will be as follows.

q = $m \times C \times \Delta T$

= $100.0 g \times 4.184 J/g^{o}C \times (27.7 - 21.5)^{o}C$

= 2594.08 J

So, heat lost by reaction = -(Heat absorbed by solution)

As, heat lost by reaction = -(2594.08 J)

therefore, heat absorbed by reaction = 2594.08 J

Now, calculate the change in enthalpy as follows.

$\Delta H = \frac{\text{Heat lost by reaction}}{\text{moles Cu(OH)_{2} formed}}$

= $\frac{-2594.08 J}{0.05 mol)}$

= -51881.6 J/mol

or, = -51.9 kJ/mol (as 1 kJ = 1000 J)

Thus, we can conclude that $\Delta H$ for the given reaction that occurs on mixing is -51.9 kJ/mol.

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