For the reaction, {\N}_2+3 {\H}_2 \to 2 {\NH}_3, If { {\d}[ {\NH}_3]}/{ {\dt}}=2 \times 10^{-4} {\mol} {\L}{ }^{-1} {\s}^{-1}, The value of {- {\d}[ {\H}_2]}/{ {\dt}} would be -

Correct Answer is : 3×10−4molL−1s−1

1 \times 10^{-4} {\mol} {\L}^{-1} {\s}^{-1}

3 \times 10^{-4} {\mol} {\L}^{-1} {\s}^{-1}

4 \times 10^{-4} {\mol} {\L}^{-1} {\s}^{-1}

6 \times 10^{-4} {\mol} {\L}^{-1} {\s}^{-1}

AIPMT 2009 Physics and Chemistry Paper

Question : 60 of 100

Marks: +1, -0

For the reaction, N2+3H2→2NH3, If ‌

d[NH3]

=2×10−4‌mol‌L‌‌−1‌s−1, The value of ‌

−d[H2]

would be -

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