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Write the equilibrium expression and equilibrium constant for the given reverse reaction. |
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Answer» Solution :`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g),K=0*50` at 6734 `K_(c)=([N_(2)][H_(2)]^(3))/([NH_(3)]^(2))`. For the reverse reaction, `K_(c)=0.50rArrK_(E)'=1/K_(e)=1/0.50=2`. |
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