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3 years ago

For the reaction PCl5(g) + heat PCl3(g) + Cl2(g), what will happen when the temperature is decreased

Chemistry

2 answers:

vovikov84 [41]3 years ago

5 0

There will be a shift towards the reactants

Rus_ich [418]3 years ago

4 0

Answer:

As temperature is decreased the reaction will shift towards the reactants i.e. it will favour the formation of PCl5

Explanation:

According to Le Chatelier's principle, when a reaction at equilibrium is subject to a change in temperature, pressure, concentration etc then the equilibrium will shift in a direction so as to undo the effect of the induced change.

The given reaction is:

PCl5(g) + heat ↔ PCl3(g) + Cl2(g)

This reaction is endothermic i.e. heat is absorbed by the reactants. When the temperature is decreased, this would imply that heat is being removed from the system. Therefore, as per Le Chatelier's principle the equilibrium will shift in a direction that would favor more input of heat i.e. it will shift towards the reactants.

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Answer:

the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

Explanation:

Given the data in the question;

$Ca^{2+$ + $y^{4-$ ⇄ $CaY^{2-$

Formation constant Kf

Kf = $CaY^{2-$ / ( [ $Ca^{2+$ ][ $y^{4-$ ] ) = 5.0 × 10¹⁰

Now,

[ $y^{4-$ ] = $\alpha _4CH_4Y$ ; ∝₄ = 0.35

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$Ca^{2+$ + $H_4Y$ ⇄ $CaY^{2-$ + 4H⁺

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so at equilibrium, $CH_4Y$ = $Ca^{2+$ = x

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since Kf is high, them x will be small so, 0.010-x is approximately 0.010

so;

Kf = $CaY^{2-$ / ( [ $Ca^{2+$ ][ $y^{4-$ ] ) = $CaY^{2-$ / ( [ $Ca^{2+$ ][ $\alpha _4CH_4Y$ ] ) = 5.0 × 10¹⁰

⇒ $CaY^{2-$ / ( [ $Ca^{2+$ ][ $\alpha _4CH_4Y$ ] ) = 5.0 × 10¹⁰

⇒ 0.010 / ( [x][ 0.35 × x] ) = 5.0 × 10¹⁰

⇒ 0.010 / 0.35x² = 5.0 × 10¹⁰

⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 5.7142857 × 10⁻¹³

⇒ x = √(5.7142857 × 10⁻¹³)

⇒ x = 7.559 × 10⁻⁷

Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

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