JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
For the decomposition of the compound, represented as \(N{H_2}COON{H_4}\left( s \right) \rightleftharpoons 2N{H_3}\left( g \right) + C{O_2}\left( g \right)\) the \(K_p\, =2.9\times10^{- 5}\, atm^3\) . If the reaction is started with \(1\, mol\) of the compound, the total pressure at equilibrium would be............\(\times10^{-2}\, atm\)
- A \(1.94\)
- B \(5.82\)
- C \(7.66\)
- D \(38.8\)
Answer & Solution
Correct Answer
(B) \(5.82\)
Step-by-step Solution
Detailed explanation
\(N{H_2}COON{H_4}(s) \leftrightarrow 2N{H_3}(g) + C{O_2}(g)\) \({K_P} = \frac{{{{({P_{N{H_3}}})}^2} \times ({P_{C{O_2}}})}}{{{P_{N{H_2}COON{H_4}(s)}}}}\) \( = {({P_{N{H_3}}})^2} \times ({P_{C{O_2}}})\) As evident by the reaction, \(NH_3\) and \(CO_2\) are formed in molar ratio…
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