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Stella [2.4K]
3 years ago
13

For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g)

Chemistry
1 answer:
aev [14]3 years ago
7 0

Answer:

ΔS° = -268.13 J/K

Explanation:

Let's consider the following balanced equation.

3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)

We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:

ΔS° = ∑np.Sp° - ∑nr.Sr°

where,

ni are the moles of reactants and products

Si are the standard molar entropies of reactants and products

ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]

ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]

ΔS° = -268.13 J/K

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The free energy of formation of nitric oxide, NO, at 1000 K (roughly the temperature in an automobile engine during ignition) is
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<u>Answer:</u> The value of K_p for the chemical equation is 8.341\times 10^{-5}

<u>Explanation:</u>

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N_2(g)+O_2(g)\rightarrow 2NO(g)

To calculate the K_p for given value of Gibbs free energy, we use the relation:

\Delta G=-RT\ln K_p

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Putting values in above equation, we get:

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Calculate the molarity of each solution.
weeeeeb [17]
Q1)
molarity is defined as the number of moles of solute in 1 L solution 
the number of moles of LiNO₃ - 0.38 mol
volume of solution - 6.14 L
since molarity is number of moles in 1 L 
number of moles in 6.14 L - 0.38 mol
therefore number of moles in 1 L - 0.38 mol / 6.14 L = 0.0619 mol/L
molarity of solution is 0.0619 M

Q2)
the mass of C₂H₆O in the solution is 72.8 g
molar mass of C₂H₆O is 46 g/mol 
number of moles = mass present / molar mass of compound
the number of moles of C₂H₆O - 72.8 g / 46 g/mol 
number of C₂H₆O moles - 1.58 mol
volume of solution - 2.34 L
number of moles in 2.34 L - 1.58 mol
therefore number of moles in 1 L - 1.58 mol / 2.34 L = 0.675 M
molarity of C₂H₆O is 0.675 M

Q3)

Mass of KI in solution - 12.87 x 10⁻³ g
molar mass - 166 g/mol
number of mole of KI = mass present / molar mass of KI
number of KI moles = 12.87 x 10⁻³ g / 166 g/mol = 0.0775 x 10⁻³ mol
volume of solution - 112.4 mL 
number of moles of KI in 112.4 mL - 0.0775 x 10⁻³ mol
therefore number of moles in 1000 mL- 0.0775 x 10⁻³ mol / 112.4 mL x 1000 mL
molarity of KI - 6.90 x 10⁻⁴ M
8 0
3 years ago
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