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`NH_(4)HS(s)hArrNH_(3)(g)+H_(2)S(g)``The3 equilibrium pressure at `25^(@)C` is 0.660 atm . What is `K_(p)` for the reaction ?A. `0.109`B. `0.218`C. `1.89`D. `2.18` |
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Answer» Correct Answer - A `NH_(4)HS(s) hArr underset(P)(NH_(3)(g)) + underset(P)(H_(2)S(g))` Equ. Pressure - 2P = `0.660` P = `0.330` `K_(p) = P_(NH_(3)) . P_(H_(2)S)` = `0.33 xx 0.33 = 0.109` |
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