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2 years ago

Consider the reaction.

Chemistry

1 answer:

TiliK225 [7]2 years ago

3 0

Answer:

0.00227 atm is the pressure of B(g)at equilibrium.

Explanation:

$2A(g)\rightleftharpoons B(g)$

The value of equilibrium constant = $K_p=7.80\times 10^{-5}$

Partial pressure of A before heating = 5.40 atm

$2A(g)\rightleftharpoons B(g)$

Initially

5.40 atm 0

at equilibrium

(5.40-2p)atm p

The expression of an equilibrium constant can be given as

$K_p=\frac{p}{(5.40-2p)^2}$

$7.80\times 10^{-5}=\frac{p}{(5.40-2p)^2}$

Solving for p ;

p = 0.00227 atm

Partial pressure of B at 500 K = 0.00227 atm

0.00227 atm is the pressure of B(g)at equilibrium.

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2 years ago

At 700 K, the reaction 2SO2(g) + O2(g) <====> 2SO3(g) has the equilibrium constant Kc = 4.3 x 106. At a certain instant, f

nadya68 [22]

Answer:

The system is not in equilibrium and will evolve left to right to reach equilibrium.

Explanation:

The reaction quotient Qc is defined for a generic reaction:

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$Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where the concentrations are not those of equilibrium, but other given concentrations

Chemical Equilibrium is the state in which the direct and indirect reaction have the same speed and is represented by a constant Kc, which for a generic reaction as shown above, is defined:

$Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where the concentrations are those of equilibrium.

This constant is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

Comparing Qc with Kc allows to find out the status and evolution of the system:

If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.

If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

In this case:

$Q=\frac{[So_{3}] ^{2} }{[SO_{2} ]^{2}* [O_{2}] }$

$Q=\frac{10^{2} }{0.10^{2} *0.10}$

Q=100,000

100,000 < 4,300,000 (4.3*10⁶)

Q < Kc

<u><em> The system is not in equilibrium and will evolve left to right to reach equilibrium.</em></u>

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2 years ago

Why is the hydrogen molecule stable with only 2 electrons while the fluorine molecule needs to have 8 electrons available for ea

Whitepunk [10]

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Explanation:

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1 year ago

A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The

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the equation we use is, $q=mc\Delta T$ .