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  •  For the decomposition of the compound,represented as NH2COONH4(s) <img alt="\rightleftharpoons" src="https://learn.careers360.com/latex-image/?%5Cdpi%7B100%7D%20%5Cright

 For the decomposition of the compound,represented as NH2COONH4(s) \rightleftharpoons 2NH3(g) + CO2(g) the Kp = 2.9 x 10-5 atm3. If the reaction is started with 1 mol of the compound, the total pressure at equilibrium would be :

Option: 1

 1.94 x 10-2 atm

 


Option: 2

 5.82 x 10-2 atm


Option: 3

 7.66 x 10-2 atm


Option: 4

 38.8 x 10-2 atm


Answers (1)

best_answer

NH_{2}COONH_{4}\left ( s \right )\rightleftharpoons 2NH_{3}\left ( g \right )+CO_{2}\left ( g \right )
     1\:mol                                        0                   0

equi.  1-\alpha                                   2\alpha               \alpha

Let \alpha = partial pressure of CO_{2}

Kp=\left ( 2\alpha \right )^{2}\left ( \alpha \right )^{1}= 4\alpha ^{3}

\Rightarrow 2.9\times 10^{-5}\:atm^{3}=4\alpha ^{3}

\Rightarrow \alpha ^{3}=7.5\times 10^{-6}

\alpha =\left ( 7.5\times 10^{-6} \right )^{\frac{1}{3}}

= 1.94\times 10^{-2}\:atm

Total\:pressure=3\alpha =5.82\times 10^{-2}\:atm

 

 

Posted by

Ajit Kumar Dubey

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