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For the decomposition reaction `NH_(2)COONH_(4)(s) hArr2NH_(3)(g)+CO_(2)(g)`, the `K_(p)=3.2xx10^(-5)atm^(3)`. The total pressure of gases at equilibrium when `1.0` mol of `NH_(2)COONH_(4)(s)` was taken to start with will be:A. `0.25atm`B. `0.12atm`C. `0.04atm`D. `0.06atm` |
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Answer» Correct Answer - D We have `NH_(2)COONH_(4)(S)hArr2NH_(3)(g)+CO_(2)(g)hArr2NH_(3)(g)+CO_(2)(g)` `K_(p)+(2p)^(2)p=4p^(3)=3.2xx10^(-5)atm^(3)` This gives `p=2xx10^(-2)atm` `p_(total)=2p+p=3p=6xx10^(-2)atm` |
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