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3 years ago

Calculate ΔSrxn∘ for the reaction

2 answers:

7 0

The entropy of the reaction can be calculated alike the enthalpy of the reaction which is equal to the difference between the summation of entropies of the products multiplied by their corresponding stoich coeff. and the summation of the entropies of the reactants multiplied by their corresponding stoich coeff. In this case, the answer is -146.8 J / mol K

Effectus [21]3 years ago

6 0

Answer:

ΔS°rxn = -146.8 J/K

Explanation:

Let's consider the following reaction.

2 NO(g) + O₂(g) → 2 NO₂(g)

The standard entropy change of the reaction (ΔS°rxn) is equal to the standard entropies of the products times their stoichiometric coefficients minus the standard entropies of the reactants times their stoichiometric coefficients.

ΔS°rxn = 2 mol × S°(NO₂(g)) - 2 mol × S°(NO(g)) - 1 mol × S°(O₂(g))

ΔS°rxn = 2 mol × 240.0 J/mol.K - 2 mol × 210.8 J/mol.K - 1 mol × 205.2 J/mol.K

ΔS°rxn = -146.8 J/K

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2 years ago

Answer the following questions based on the reaction below: NaOH(aq) + KHP(s) --> NaKP(aq)+H2O(I)

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a) 1.44g

b) 77.3%

Explanation:

1. Chemical balanced equation (given)

$NaOH(aq)+ KHP(s)\rightarrow NaKP(aq)+H_2O(l)$

2. Mole ratio

$1molNaOH(aq):1molKHP(s)$

This is, 1 mol of NaOH will reacts with 1 mol of KHP.

3. Find the number of moles in 72.14 mL of the base

$Molarity=\text{number of moles of solute}/\text{volume of solution in liters}$

$\text{Volume of solution}=72.14mL=0.07214liters$

$\text{Number of moles of NaOH}=0.0978M\times 0.07214liter=0.007055mol$

4. Find the number of grams of KHP that reacted

The number of moles of KHP that reacted is equal to the number of moles of NaOH, 0.007055 mol

Convert moles to grams:

mass = number moles × molar mass = 0.007055mol × 204.23g/mol

mass = 1.4408 g.

You have to round to 3 significant figures: 1.44 g (because the molarity is given with 3 significant figures).

5. Find the percentage of KHP in the sample

The percentage is how much of the substance is in 100 parts of the sample.

The formula is:

% = (mass of substance / mass of sample) × 100

% = (1.4408g/ 1.864g) × 100 = 77.3%

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3 years ago

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