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Andru [333]
3 years ago
14

What is the correct equilibrium constant expression for equation P2(g)

Chemistry
1 answer:
sertanlavr [38]3 years ago
6 0

Answer:

k = [F2]² [PO]² / [P2] [F2O]²

Explanation:

In a chemical equilibrium, the equilibrium constant expression is written as the ratio between the molar concentration of the products over the molar concentration of the reactants. Each species powered to its reaction coefficient. For the equilibrium:

P2(g) + 2F2O(g) ⇄ 2PO(g) + 2F2(g)

The equilibrium constant, k, is:

k = [F2]² [PO]² / [P2] [F2O]²

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Explanation:

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A __________ compound is a cyclic compound in which one or more of the ring atoms is an atom other than carbon. g
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At what time will the pressure of SO₂Cl₂ decline to 0.50 its initial value? Express your answer using two significant figures.
QveST [7]

Answer: The time is 0.69/k seconds

Explanation:

The following integrated first order rate law

ln[SO₂Cl₂] - ln[SO₂Cl₂]₀ = - k×t

where

[SO₂Cl₂] concentration at time t,

[SO₂Cl₂]₀ initial concentration,

k rate constant

Therefore, the time elapsed after a certain concentration variation is given by:

t=\frac{ln[SO_{2}Cl_{2}]_{0} - ln[SO_{2}Cl_{2}]}{k}=\frac{ln\frac{[SO_{2}Cl_{2}]_{0}}{[SO_{2}Cl_{2}]} }{k}

We could assume that SO₂Cl₂ behaves as a ideal gas mixture so partial pressure is proportional to concentration:

p_{(SO_{2}Cl_{2})}V = n_{(SO_{2}Cl_{2})}RT

[SO_{2}Cl_{2}]= \frac{n_{(SO_{2}Cl_{2})}}{V}}=\frac{p_{(SO_{2}Cl_{2})}}{RT}}

In conclusion,

t = ln( p(SO₂Cl₂)₀/p(SO₂Cl₂) )/k

t=\frac{ln\frac{p_{(SO_{2}Cl_{2})}_{0}}{p_{(SO_{2}Cl_{2})}} }{k}

for

p_{(SO_{2}Cl_{2})}=0.5p_{(SO_{2}Cl_{2})}_{0}

t=\frac{ln\frac{p_{(SO_{2}Cl_{2})}_{0}}{0.5p_{(SO_{2}Cl_{2})_{0}}} }{k}

t=\frac{ln\frac{1}{0.5} }{k}

t=\frac{ln(2)}{k}

t=\frac{0.69}{k}}

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