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The equilibrium constant of the reaction...

The equilibrium constant of the reaction `A_(2)(g)+B_(2)(g)hArr2AB(g)` at `100^(@)C` is 50. If a one litre flask containing one mole of `A_(2)` is connected to a two litre flask containing two moles of `B_(2)`, how many moles of `AB` will be formed at `373 K`?

Text Solution

Verified by Experts

` {: (,A_(2),+,B_(2),hArr,2AB),(" Intial amounts:",1 "mole",,2 "mole",,0),("Amounts at eqm:",1-x,,2-x,,2x),(" Molar concs at eqm.",(1-x)/3,,(2-x)/3,,(2x)/3 "mol"L^(1)):}`
` K = ([AB]^(2))/([A_(2)][B_(2)]) or 50 = (2x^(2))/(((1-x)/3)((2-x)/3))`
On solving , we get `x= 0*955 ` mole
Hence, no. of moles of AB formed at eqm. `= 2 xx 0* 955 = 1*91 ` moles
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The equilibrium constant for the reaction A_(2)(g) +B_(2) hArr 2AB(g) at 373 K is 50. If one litre flask containing one mole of A_(2) is connected to a two flask containing two moles of B_(2) , how many moles of AB will be formed at 373 K?

Knowledge Check

  • The equilibrium constant of the reaction A_(2) (g) + B_(2) (g) hArr 2AB (g) at 373 K is 50. If 1 L of flask containing 1 mole of A_(2) (g) is connected to 2 L flask containing 2 moles B_(2) (g) at 100^(@)C , the amount of AB produced at equilibrium at 100^(@)C would be

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    C
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    D
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