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ryzh [129]
3 years ago
10

Which of the following statements explains which trial has a lower concentration of the reactant? (5 points) Trial 1, because th

e average rate of the reaction is lower. Trial 1, because this reaction lasted for a longer duration than Trial 2. Trial 2, because this reaction was initially fast and later slowed down. Trial 2, because the v
Chemistry
2 answers:
Alla [95]3 years ago
8 0

Answer:

A)Trial 1 because the average rate of reaction is lower.

Explanation:

JuSt ToOk ThE tEsT

PIT_PIT [208]3 years ago
5 0

Answer:

A)Trial 1 because the average rate of reaction is lower.

Explanation:

I accidentally gave myself low rating my bad

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Consider the reaction:
goldfiish [28.3K]

Answer:

K = Ka/Kb

Explanation:

P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?

P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka

PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb

K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)

Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)

Kb = [PCl₅]/ ([PCl₃] [Cl₂])

Since [PCl₅] = [PCl₅]

From the Ka equation,

[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)

From the Kb equation

[PCl₅] = Kb ([PCl₃] [Cl₂])

Equating them

Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])

(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)

(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)

Comparing this with the equation for the overall equilibrium constant

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