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In a closed vessel of 1 dm^(3) capacity,...

In a closed vessel of 1 `dm^(3)` capacity, 1 mol `N_(2)(g)` and 2 mol `H_(2)(g)` interact to produce 0.8 mol `NH_(3)(g)` in the equilibrium mixture. What is the concentration of `H_(2)(g)` in the equilibrium mixture?

Text Solution

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Equilibrium of the equilibrium reaction:
`N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g)`
It is observed from the reaction that 1 mol `N_(2)(g) and 3 " mol "H_(2)(g)` are necessary for the production of 2 mol `NH_(3)(g)`, the number of moles of `H_(2)(g)` required`=(3)/(2)xx0.8=1.2` mol.
Hence, the number of moles of `H_(2)(g)` remaining in the equilibrium mixture `=2-.12=0.8` and its molar concentration`=0.8mol*L^(-1)" "[because 1 dm^(3)=1L]`.
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