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Wittaler [7]
3 years ago
15

What is Loschmidt’s number? How is it related to Avogadro’s number?

Chemistry
1 answer:
maxonik [38]3 years ago
5 0
1. Loschmidt’s number also called as Loschmidt’s constant, n₀ is the number of particles of an ideal gas in a given volume (m³). The  unit is m⁻³. At standard pressure (1 atm) and temperature (0 °C) the value of <span>Loschmidt’s number is 2.686 x 10</span>²⁵ m⁻³.

2. 
Loschmidt’s number,n₀ can be expressed as,
        n₀ =  \frac{P}{kT}
where P is the pressure of the gas, K is the Boltzmann constant and T is the <span>thermodynamic temperature.

Since, k = R/NA (where R is the universal gas constant and NA is the Avogadro constant)
</span>n₀ = \frac{P}{(R/NA)T}
n₀ = \frac{PNA}{RT}

Hence, n₀ α NA

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At a certain temperature, 0.760 mol SO3 is placed in a 1.50 L container. 2SO3(g) = 2SO2(g) + O2(g)At equilibrium, 0.130 mol O2 i
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Molarity=\frac{0.760}{1.50\ L}

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Considering the ICE table for the equilibrium as:

\begin{matrix} & 2SO_3_{(g)} & \rightleftharpoons & 2SO_2_{(g)} & + & O_2_{(g)}\\At\ time, t = 0 & 0.5067 &&0&&0 \\ At\ time, t=t_{eq} & -2x &&2x&&x \\----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:- &0.5067-2x&&2x&&x\end{matrix}

Given:    

Equilibrium concentration of  O₂ = 0.130 mol

Volume = 1.50 L

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity=\frac{0.130}{1.50\ L}

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[SO₃] = 0.5067-2x = 0.3334 M

The expression for the equilibrium constant is:

K_c=\frac {[SO_2]^2[O_2]}{[SO_3]^2}  

K_c=\frac{(0.3334)^2\times 0.0867}{(0.3334)^2}

K_c=0.0867

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