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timama [110]
4 years ago
13

A liter of the gas mixture at equilibrium contains 0.0045 mol of N2O4 and 0.030 mol of NO2 at 10C. Write the expression for the

equilibrium constant (Keq) and calculate the value of the constant for the reaction.
Chemistry
1 answer:
Alona [7]4 years ago
7 0

Answer:

Keq = [NO₂]²/[N₂O₄] = 0.2.

Explanation:

  • For the gas mixture equilibrium:

<em>N₂O₄ ⇄ 2NO₂</em>

The expression of the equilibrium constant (Keq):

Keq = [NO₂]²/[N₂O₄]

<em>∴ Keq = [NO₂]²/[N₂O₄]</em> = (0.03)²/(0.0045) = <em>0.2.</em>

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7 0
3 years ago
A sample of gas contains 0.1700 mol of NH3(g) and 0.2125 mol of O2(g) and occupies a volume of 17.8 L. The following reaction ta
telo118 [61]

Answer:

The volume of the sample after the reaction takes place is 19.78 L.

Explanation:

The given variables are;

Number of moles of NH₃(g) = 0.1700 mol

Number of moles of O₂(g) = 0.2125 mol

Volume occupied by the mixture = 17.8 L

The reaction

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

Then takes place

That is 4 moles of NH₃(g) reacts with 5 moles of O₂(g) to produce 4 moles of NO(g) and 6 moles of H₂O(g).

Since there are less number of moles of NH₃(g) (= 0.1700 mol) in the mixture, we factor the above equation by the number of moles of NH₃(g)  present.

That is,

1 moles of NH₃(g) reacts with 5/4 moles of O₂(g) to produce 1 moles of NO(g) and 3/2 moles of H₂O(g).

Therefore,

0.1700 mol of NH₃(g) reacts with 5/4×0.1700  moles of O₂(g) to produce 0.1700  moles of NO(g) and 3/2×0.1700  moles of H₂O(g).

Which gives

0.1700 mol of NH₃(g) reacts with 0.2125  moles of O₂(g) to produce 0.1700  moles of NO(g) and 0.255  moles of H₂O(g).

Therefore, all of the NH₃(g) and O₂(g)  are consumed in the reaction and the present gases in sample then becomes

0.1700  moles of NO(g) and 0.255  moles of H₂O(g).

Total number of moles of reactant = 0.17 + 0.2125 = 0.3825

Total number of moles of product formed = 0.17 + 0.255 = 0.425

However, Avogadro's law states that equal volume of all gases at the same temperature and pressure contains equal number of molecules.

That is volume occupied by  0.3825 moles of gas = 17.8 L

Therefore the volume occupied by  0.425 moles of gas = 17.8×0.425/0.3825 L = 19.78 L

3 0
4 years ago
What letter on the model titration curve corresponds to the point where ph equals the numerical value of pka for hpr? what speci
irakobra [83]

Letter C on the model titration curve corresponds to the point where pH equals the numerical value of pKa for HPr

<h3>What is a titration curve?</h3>

A titration curve is a graph of the pH of a solution against increasing volumes of an acid or a base that is added to the solution.

The pH of a solution is the negative logarithm to base ten of the hydrogen ion concentration and is a measure of the acidity or alkalinity of the solution.

The pKa is the acid dissociation constant of an acid solution.

In a titration of a strong acid and strong base, the pH at equivalence point is equal to the pKa of the acid.

The equivalence point is the point when equal moles of acids and base has reacted.

In the given titration curve, pH = pKa at point C.

In conclusion, for a titration curve of strong acid and base, at equivalence point, pH is equal to pKa of acid.

Learn more about equivalence point at: brainly.com/question/23502649

#SPJ1

4 0
2 years ago
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