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sergij07 [2.7K]
3 years ago
6

9. At equilibrium a 2 L vessel contains 0.360

Chemistry
1 answer:
klio [65]3 years ago
3 0

Answer:

Ke = 34570.707

Explanation:

  • H2(g) + Br2(g) → 2 HBr(g)

equilibrium constant (Ke):

⇒ Ke = [HBr]² / [Br2] [H2]

∴ [HBr] = (37.0 mol) / (2 L) = 18.5 mol/L

∴ [Br2] = (0.110 mol) / (2 L) = 0.055 mol/L

∴ [H2] = (0.360 mol) / (2 L) = 0.18 mol/L

⇒ Ke = (18.5 mol/L)² / (0.055 mol/L)(0.18 mol/L)

⇒ Ke = 34570.707

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If a system has a reaction quotient of 2.13 ✕ 10−15 at 100°C, what will happen to the concentrations of COBr2, CO, and Br2 as th
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This is an incomplete question, here is a complete question.

Consider the following equilibrium at 100°C.

COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)

K_c=4.74\times 10^4

Concentration at equilibrium:

[COBr_2]=1.58\times 10^{-6}M

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If a system has a reaction quotient of 2.13 × 10⁻¹⁵ at 100°c, what will happen to the concentrations of COBr₂, Co and Br₂ as the reaction proceeds to equilibrium?

Answer : The concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

Explanation :

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)

The expression for reaction quotient will be :

Q=\frac{[CO][Br_2]}{[COBr_2]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(2.78\times 10^{-3})\times (2.51\times 10^{-5})}{(1.58\times 10^{-6})}=4.42\times 10^{-2}

The given equilibrium constant value is, K_c=4.74\times 10^4

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When Q>K_c that means product > reactant. So, the reaction is reactant favored.

When Q that means reactant > product. So, the reaction is product favored.

When Q=K_c that means product = reactant. So, the reaction is in equilibrium.

From the above we conclude that, the Q that means product < reactant. So, the reaction is product favored that means reaction must shift to the product (right) to be in equilibrium.

Hence, the concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

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