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Chemical Equilibrium Chemistry Class 11th Equilibrium

At 1990K and 1 atm pressure, there are equal number of Cl₂ molecules and Cl atoms in the reaction mixture. The value of K_P for the reaction Cl_2(g) $?$ 2Cl_(g) under the above conditions is x × 10^-1. The value of x is______ (Rounded off to the nearest integer)

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alok kumar singh | Contributor-Level 10

4 weeks ago

$C l_{2 (g)} ? 2 C l_{(g)}$

Initially -> 1 mol -

At eq. 1-x mol 2x mol

Here; molecules of Cl₂ = atoms of Cl

i.e moles of Cl₂ = moles of Cl

So : 1 – x = 2x x = 1/3

Moles of Cl₂ at

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$C l_{2 (g)} ? 2 C l_{(g)}$

Initially -> 1 mol -

At eq. 1-x mol 2x mol

Here; molecules of Cl₂ = atoms of Cl

i.e moles of Cl₂ = moles of Cl

So : 1 – x = 2x x = 1/3

Moles of Cl₂ at equibrium = $\frac{2}{3}$

Moles of Cl at equilibrium = $\frac{2}{3}$

Total moles = $\frac{4}{3}$

Now: $P_{C l_{2}} = \frac{2}{4} \times 1 a t m$

$= \frac{1}{2} a t m$

$P_{C l} = \frac{2}{4} \times 1 a t m = \frac{1}{2} a t m$

$K_{P} = \frac{{(P_{C l})}^{2}}{P_{C l_{2}}} = \frac{{(\frac{1}{2})}^{2}}{\frac{1}{2}} = \frac{1}{2}$

$K_{P} = 0.5 = 5 \times 1 0^{- 1}$

Here : x = 5

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At 1990K and 1 atm pressure, there are equal number of Cl₂ molecules and Cl atoms in the reaction mixture. The value of K_P for the reaction Cl_2(g) $?$ 2Cl_(g) under the above conditions is x × 10^-1. The value of x is______ (Rounded off to the nearest integer)

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At 1990K and 1 atm pressure, there are equal number of Cl2 molecules and Cl atoms in the reaction mixture. The value of KP for the reaction Cl2(g) ? 2Cl(g) under the above conditions is x × 10-1. The value of x is______ (Rounded off to the nearest integer)

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At 1990K and 1 atm pressure, there are equal number of Cl₂ molecules and Cl atoms in the reaction mixture. The value of K_P for the reaction Cl_2(g) $?$ 2Cl_(g) under the above conditions is x × 10^-1. The value of x is______ (Rounded off to the nearest integer)