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Introduction to Oxidation and Reduction

The n factor ...

Question

The n factor for ${N H}_{3}$ in the reaction:
$N_{2} + {3 H}_{2} ⟶ {2 N H}_{3}$

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Solution

The correct option is C 1
n-factor is the number of moles of electrons given or gained.
$N H_{3}$ accepts $1 H^{+} i o n$ So its N-factor is 1

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Similar questions

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

K \frac{{[N H_{3}]}_{3}^{2}}{[N_{2}] {[H_{2}]}_{2}^{3}}

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

K = \frac{[N H_{3}]^{2}}{[N_{2}] [H_{2}]^{3}}

N_{2}

N H_{3}

N_{2} + 3 H_{2} \to 2 N H_{3}

2 N H_{3} \to N_{2} + 3 H_{2}

- \frac{Δ [N H_{3}]}{Δ t} = K_{1} [N H_{3}]; \frac{Δ [N_{2}]}{Δ t} = K_{2} [N H_{3}]; \frac{Δ [H_{2}]}{Δ t} = K_{3} [N H_{3}]

k_{1}, k_{2}

k_{3}

N_{2} + 3 H_{2} ⟶ 2 N H_{3}

E_{1}

E_{2}

N H_{3}

N_{2}

E_{1} - E_{2}

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