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2 years ago

At a certain temperature this reaction follows first-order kinetics with a rate constant of 0.086 s^-1:

1 answer:

8 0

It takes 31 s for 1.27 M H₃PO₄ to decrease its concentration to 7.0% of its initial value following first-order kinetics.

<h3>What is first-order kinetics?</h3>

First-order kinetics occur when a constant proportion of a reactant disappears per unit time.

Let's consider the following first-order kinetics reaction with a rate constant k = 0.086 s⁻¹.

2 H₃PO₄(aq) = P₂O₅(aq) + 3 H₂O(aq)

Given the initial concentation is [H₃PO₄]₀ = 1.27 M, the concentration representing 7.0% of this value is:

[H₃PO₄] = 7.0% × 1.27 M = 0.089 M

We can calculate the time elapsed (t) using the following expression.

ln ([H₃PO₄]/[H₃PO₄]₀) = - k × t

ln (0.089 M/1.27 M) = - 0.086 s⁻¹ × t

t = 31 s

It takes 31 s for 1.27 M H₃PO₄ to decrease its concentration to 7.0% of its initial value following first-order kinetics.

Learn more about first-order kinetics here: brainly.com/question/18916637

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1.33 dm3 of water at 70°C are saturated by 2.25

astraxan [27]

Given that 4.50 dm³ of Pb(NO₃)₂ is cooled from 70 °C to 18 °C, the

amount amount of solute that will be deposited is 1,927.413 grams.

<h3>How can the amount of solute deposited be found?</h3>

The volume of water 1.33 dm³ of water 70 °C.

The number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water at 70 °C = 2.25 moles

At 18 °C, the number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water = 0.53 moles

Therefore;

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C is therefore;

1.33 dm³ contains 2.25 moles.

$Number \ of \ moles \ in \ 4.50 \ dm^3 = \dfrac{2.25}{1.33} \times 4.50 \approx \mathbf{7.613 \, moles}$

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C ≈ 7.613 moles

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C is therefore;

1.33 dm³ contains 0.53 moles

$Number \ of \ moles \ in \ 4.50 \ dm^3 = \dfrac{0.53}{1.33} \times 4.50 \approx \mathbf{1.79 \, moles}$

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C ≈ 1.79 moles

The number of moles that precipitate out = The amount of solute deposited

Which gives;

Amount of solute deposited = 7.613 moles - 1.79 moles = 5.823 moles

The molar mass of Pb(NO₃)₂ = 207 g + 2 × (14 g + 3 × 16 g) = 331 g

The molar mass of Pb(NO₃)₂ = 331 g/mol

The amount of solute deposited = Number of moles × Molar mass

Which gives;

The amount of solute deposited = 5.823 moles × 331 g/mol =<u> 1,927.413 g </u>

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