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Standard Electrode Potentials

Which of the ...

Question

Which of the following statement(s) is/are correct about the rate expression? $(K_{1}$ and $K_{2}$ are reaction constant for forward and backward reaction respectively. $)$

\frac{d_{[} u r e a]}{d t} = \frac{k_{1} k_{3}}{k_{2}} [N H_{4} N C O]

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\frac{d_{[} u r e a]}{d t} = \frac{k_{1} k_{3}}{k_{2} K_{4}} [N H_{4} N C O]

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\frac{d_{[} u r e a]}{d t} = k [N H_{4} N C O]

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\frac{d_{[} u r e a]}{d t} = \frac{k_{1} \times k_{2}}{k_{3} \times k_{4}} [N H_{4} N C O]

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Solution

The correct options are
A $\frac{d_{[} u r e a]}{d t} = \frac{k_{1} k_{3}}{k_{2}} [N H_{4} N C O]$
C $\frac{d_{[} u r e a]}{d t} = k [N H_{4} N C O]$
$\frac{d}{d t} [N H_{2} C O M H_{2}] = k_{4} [N H_{3}] [H N C O]$ from (iii)
Applying steady state approximation to $H N C O$ or $N H_{3}$

$\frac{d [H N C O]}{d t} = O = k_{3} [N H_{4} N C O] k_{4} [N H_{3}] [H N C O]$
$∴ \frac{k_{3}}{k_{4}} = \frac{[N H_{3}] [H N C O]}{[N H_{4} C N O]}$
$\frac{d [u r e a]}{d t} = k_{4} \times [N H_{3}] [H N C O] = k_{4} \times \frac{k_{3}}{k_{4}} [N H_{4} N C O]$
Also, $[N H_{4} N C O] = \frac{k_{1}}{k_{2}} [N H_{4} C N O]$
$∴ \frac{d [u r e a]}{d t} = k_{3} \times \frac{k_{1}}{k_{2}} \times [N H_{4} C N O]$
$= k [N H_{4} C N O]$

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Similar questions

2 N O (g) + 2 H_{2} (g) \to N_{2} (g) + 2 H_{2} O (g)

\frac{d [N_{2}]}{d t} = k_{1} [N O] [H_{2}]; \frac{d [H_{2} O]}{d t} = k_{2} [N O] [H_{2}]; \frac{- d [N O]}{d t} = k_{3} [N O] [H_{2}]; \frac{d [H_{2}]}{d t} = k_{4} [N O] [N_{2}]

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N_{2} O_{5} ⟶ 2 N O_{2} + \frac{1}{2} O_{2}

\frac{- d [N_{2} O_{5}]}{d t} = k_{1} [N_{2} O_{5}]

\frac{d [N O_{2}]}{d t} = k_{2} [N_{2} O_{5}]

\frac{d [O_{2}]}{d t} = k_{3} [N_{2} O_{5}]

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- \frac{d [A]}{d t} = k_{1} {[A]}^{2}; \frac{d [B]}{d t} = k_{2} {[B]}^{2}; \frac{d [A]}{d t} = k_{3} {[A]}^{2}

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2 N_{2} O_{5} \to 4 N O_{2} + O_{2}

- \frac{d [N_{2} O_{5}]}{d t} = k_{1} [N_{2} O_{5}]

\frac{d [N O_{2}]}{d t} = k_{2} [N_{2} O_{5}]

\frac{d [O_{2}]}{d t} = k_{3} [N_{2} O_{5}]

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