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2 years ago

Dissolution of KOH, ΔHsoln:

Chemistry

1 answer:

swat322 years ago

6 0

Using Hess's law we found:

1) By adding reaction 10.2 with the reverse of reaction 10.1 we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) ΔH (10.3)

2) The ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy (ΔH).

The reactions of dissolution (10.1) and neutralization (10.2) are:

KOH(s) → KOH(aq) ΔHsoln (10.1)

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq) ΔHneut (10.2)

1) According to Hess's law, the total change in enthalpy of a reaction resulting from differents changes in various reactions can be calculated as the sum of all the enthalpies of all those reactions.

Hence, to get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

We need to add reaction 10.2 to the reverse of reaction 10.1

KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)

Canceling the KOH(s) from both sides, we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

2) The change in enthalpy for reaction 10.3 can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:

$\Delta H = \Delta H_{soln} + \Delta H_{neut}$

The enthalpy of reaction 10.1 (ΔHsoln) changed its sign when we reversed reaction 10.1, so:

$\Delta H = \Delta H_{neut} - \Delta H_{soln}$

Therefore, the ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy ΔH.

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4 years ago

The combustion of 40.10 g of a compound which contains only C, H, Cl and O yields 58.57 g of CO2 and 14.98 g of H2O. Another sam

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The empirical formula is C4H5ClO2

Explanation:

Step 1: Data given

Mass of the sample = 40.10 grams

Mass of CO2 produced = 58.57 grams

Mass of H2O produced = 14.98 grams

Molar mass CO2 = 44.01 g/mol

Molar mass H2O = 18.02 g/mol

Atomic mass C= 12.01 g/mol

Atomic mass O = 16.0 g/mol

Atomic mass H = 1.01 g/mol

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Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 58.57 grams / 44.01 g/mol

Moles CO2 = 1.33 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol CO2

For 1.33 moles CO2 we have 1.33 moles C

Step 4: Calculate mass C

Mass C = 1.33 grams * 12.01 g/mol

Mass C = 15.97 grams

Step 5: Calculate moles H2O

Moles H2O= 14.98 grams /18.02 g/mol

Moles H2O = 0.831 moles

Step 6: Calculate moles H

For 1mol H2O we have 2 moles H

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Mass H = 1.662 moles * 1.01 g/mol

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Step 8: Calculate mass %

%C = (15.97 grams / 40.10) * 100 %

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%Cl = (22.06 / 75.00 ) * 100%

%Cl = 29.4 %

%O = 100 % - 39.8% - 4.2 % - 29.4 %

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Step 9: Calculate moles in compound

We assume the compound has a mass of 100 grams

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MAss Cl = 29.4 grams

Mass O = 26.6 grams

Moles C = 39.8 grams / 12.01 g/mol

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Moles H = 4.2 moles / 1.01 g/mol

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Moles Cl =29.4 grams / 35.45 g/mol

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Moles O = 26.6 grams / 16.0 g/mol

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Step 10: calculate the mol ratio

We divide by the smallest amount of moles

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H: 4.158 moles / 0.829 moles = 5

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O: 1.663 moles / 0.829 moles = 2

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