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Ammonium carbamate decomposes as `NH_(2)COONH_(4)(s) hArr 2NH_(3)(g) +CO_(2)(g)` . The value of `K_(p)` for the reaction is `2.9 xx 10^(-5) atm^(3)`. If we start the reaction with 1 mole of the compound, the total pressure at equilibrium would beA. 0.0766 atmB. 0.0194 atmC. 0.194 atmD. 0.0582 atm |
Answer» Correct Answer - D `NH_(2)COONH_(4)(s) hArR 2NH_(3)(g)+CO_(2)(g)` If partial pressure of `CO_(2)` at equilibrium is p Then partial pressure of `NH_(3)` is 2p `K_(p)=(p_(NH_(3)))^(2)xx(p_(CO_(2)))=(2p)^(2)(p)=4p^(3)` Now `4p^(3)=2.9 xx 10^(-5) `or `p=1.935xx10^(-2)` `=5.81 xx 10^(-2) atm ` |
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