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Consider the decomposition of solid `NH_(4)HS` in a flask containing `NH_(3)(g)` at a pressure of `2` atm. What will be the partial pressure of `NH_(3)(g) "and" H_(2)S(g)` after the equilibrium has been attained? `K_(p)` for the reaction is `3`.A. `P_(NH_(3))=4 "atm", P_(H_(2))s=2` atmB. `P_(MH_(3))=1.732 "atm", P_(h_2)s=1.732` atmC. `P_(NH_(3))=3 "atm", P_(H_(2))s=1` atmD. `P_(NH_(3))=1 "atm", P_(H_(2))s=1` atm |
Answer» Correct Answer - C | |