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1 year ago

Explain why doesn’t the total pressure increase when more gas is added to the chamber?

1 answer:

german1 year ago

4 0

Pressure is inversely proportional to volume. Therefore, the effect of pressure change is opposite to the effect of volume change. So when more gas is added to the chamber the total pressure of the chamber doesn't increase.

<h3>What are the different relations between pressure and volume?</h3>

As the volume changes, the concentrations and partial pressures of both reactants and products change.
As the volume decreases, the reaction shifts to the reaction side with fewer gas particles.
As the volume increases, the reaction shifts to the side of the reaction containing more gas particles.
As the pressure increases, the equilibrium shifts towards reactions with fewer moles of gas.
As the pressure decreases, the equilibrium shifts to the side of the reaction with higher moles of gas.
Moreover, the pressure change in the system due to the addition of the inert gas is not limited to this.

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Answer : The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

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b = order with respect to B

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$3.20\times 10^{-1}=k(1.50)^a(1.50)^b$ ....(1)

Expression for rate law for second observation:

$3.20\times 10^{-1}=k(1.50)^a(2.50)^b$ ....(2)

Expression for rate law for third observation:

$6.40\times 10^{-1}=k(3.00)^a(1.50)^b$ ....(3)

Dividing 1 from 2, we get:

$\frac{3.20\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(1.50)^a(2.50)^b}{k(1.50)^a(1.50)^b}\\\\1=1.66^b\\b=0$

Dividing 1 from 3, we get:

$\frac{6.40\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(3.00)^a(1.50)^b}{k(1.50)^a(1.50)^b}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[A]^1[B]^0$

$\text{Rate}=k[A]$

Thus,

The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.

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