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sergiy2304 [10]
3 years ago
5

Phosphorus pentachloride decomposes according to the chemical equation

Chemistry
1 answer:
Elena L [17]3 years ago
3 0

Answer:

[PCl₅]  → 4.47×10⁻³ M

[PCl₃]  → 0.0897M

Explanation:

We state the equilibrium: PCl₅ (g) ⇄  PCl₃(g) + Cl₂(g)

Initially we have 0.4192 moles of  PCl₅ so during the reaction, x moles of it have been reacted. In conclussion after the equilibrium we may have x moles of PCl₃ and Cl₂. In order to find the final concentration, we use the Kc, but we need molar concentrations. That's why we divide all the values with the 4.5L, volume for the vessel. The expression for Kc is:

Kc = [PCl₃] . [Cl₂] / [PCl₅]

1.80 = (x/ 4.45) . (x/ 4.45) / ((0.4192-x) /4.45)

1.80 = (x² / 4.45²) / ((0.4192-x) /4.45)

1.80 = x² / 4.45 / (0.4192-x)

1.80 (0.4192-x) = x² / 4.45

1.80 (0.4192-x) . 4.45 - x² = 0

3.36 - 8.01x - x² = 0 → Quadractic

a = -1  b = -8.01  c = 3.36

(-b + √(b² - 4ac)) / (2a)

(-(-8.01) + √((-8.01)² - 4(-1)3.36)) / (2(-1) = 0.3995

[PCl₅] = (0.4192 - 0.3995) / 4.45L → 4.47×10⁻³ M

[PCl₃] = 0.3995 /4.45L → 0.0897M

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7. There are 7. 0 ml of 0.175 M H2C2O4 , 1 ml of water , 4 ml of 3.5M KMnO4 what is the molar concentration ofH2C2O4 ?
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Answer:

7. 0.1021 M

8. 1.167 M

10. Increase in volume of water would lower the rate of reaction

Explanation:

7. What is the molar concentration of H₂C₂O₄ ?

Since we have 7.0 ml of 0.175 M H₂C₂O₄, the number of moles of H₂C₂O₄ present n = molarity of H₂C₂O₄ × volume of H₂C₂O₄ = 0.175 mol/L × 7.0 ml = 0.175 mol/L × 7 × 10⁻³ L = 1.225 × 10⁻³ mol.

Also, the total volume present V = volume of H2C2O4 + volume of water + volume of KMnO4 = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of H₂C₂O₄, M = number of moles of H₂C₂O₄/volume = n/V

= 1.225 × 10⁻³ mol/12 × 10⁻³ L

= 0.1021 mol/L

= 0.1021 M

8. Using the data from question 7 what is the molar concentration of KMnO₄ ?

Since we have 4.0 ml of 3.5 M KMnO₄, the number of moles of KMnO4 present n' = molarity of KMnO₄ × volume of KMnO₄ = 3.5 mol/L × 4.0 ml = 3.5 mol/L × 4 × 10⁻³ L = 14 × 10⁻³ mol.

Also, the total volume present V = volume of KMnO₄ + volume of water + volume of KMnO₄ = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of KMnO₄, M' = number of moles of KMnO₄/volume = n'/V

= 14 × 10⁻³ mol/12 × 10⁻³ L

= 1.167 mol/L

= 1.167 M

10. From question number 7, what effect increasing the volume of water has on the reaction rate?

Increase in volume of water would lower the rate of reaction because, the particles of both substances would have to travel farther distances to collide with each other, since there are less particles present in the solution and thus, the concentration of the particles would decrease thereby decreasing the rate of reaction.

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3 years ago
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