For a reaction 2NH3 → N2 + 3H2, it is observed that $- \frac{d\left\lbrack NH_{3} \right\rbrack}{dt}

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For a reaction 2NH3 → N2 + 3H2, it is observed that $- \frac{d\left\lbrack NH_{3} ightbrack}{dt} = K_{1}\left\lbrack NH_{3} ightbrack; + \frac{d\left\lbrack N_{2} ightbrack}{dt} = K_{2}\left\lbrack NH_{3} ightbrack	ext{ and } + \frac{d\left\lbrack H_{2} ightbrack}{dt} = K_{3}\left\lbrack NH_{3} ightbrack$The correct relation between K1, K2 and K3 is

Accepted Answer

$r_{rxn} = - \frac{1}{2}.\frac{d\left\lbrack NH_{3} \right\rbrack}{dt} = + \frac{d\left\lbrack N_{2} \right\rbrack}{dt} = + \frac{1}{3}.\frac{d\left\lbrack H_{2} \right\rbrack}{dt}$

Or, $\frac{1}{2}.K_{1}\left\lbrack NH_{3} \right\rbrack = K_{2}\left\lbrack NH_{3} \right\rbrack = \frac{1}{3}.K_{3}\left\lbrack NH_{3} \right\rbrack$

$\therefore 3K_{1} = 6K_{2} = 2K_{3}$

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