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Faraday's Second Law

N2 + 3H2 = 2...

Question

N2 + 3H2 = 2NH3 in above reaction equivalent weight of NH3 and H2 are 1) M÷3,M÷3 2) M÷2,M÷3 3)M÷3,M÷3 4)M÷2,M÷

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

C a (O H)_{2} + H_{3} P O_{4} - ------ - \to C a H P O_{4}

Z n - -- - + H_{2} S O_{4} \to Z n S O_{4} + H_{2}

2 F e S O_{4} - ------- - + H_{2} S O_{4} + O \to F e (S O_{4})_{3} + H_{2} O

\frac{M}{6}

\frac{M}{3}

\frac{M}{2}

\frac{M}{2}

M

= M

N_{2} + 3 H_{2} ⇌ 2 N H_{3}, K_{6}

2.37 \times 10^{- 3}

[N_{2}] = 2. M [H_{2}] = 3 M

N H_{3}

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

N_{2}, H_{2}

N H_{3}

1 : 3 : 2

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}}

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