1.

28 g N2 and 6.0 g of H_(2)are heated over catalyst in a closed one litre flask of 450 °C. The entire equilibrium mixture required 500 mL of 1.0 M H_(2)SO_(4)for neutralisation. Calculate the value of Kc in L^(2) "mol"^(2)for the given reaction. N_(2)(g) + 3H_(2)(g) to 2NH_(3)(g)

Answer»


SOLUTION :MOLES of `N_(2) =28/28=1,` moles of `H_(2)=6/2=3`
Moles of `H_(2)SO_(4)` required `=(500 xx 1)/1000 =0.5`
Moles of `NH_(3)` neutralised by `H_(2)SO_(4)=1.0`
`2NH_(3) + H_(2)SO_(4) to (NH_(4))_(2)SO_(4)`
Hence 1 mole of `NH_(3)` by the reaction between `N_(2)` and `H_(2)`
`{:(N_(2),+,3H_(2), `K_( c) =(1 xx 1)/(0.5 xx (1.5)^(3)) = 0.592 mol^(-2) L^(2)`


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