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JEE Main levelSingle correct optionDifficulty 2 of 3ck190

In $4 NH X_{3} + 5 O X_{2} 4 NO + 6 H X_{2} O$ , ammonia is consumed at $0.24 M s^{- 1}$ . The rate of formation of water, in $M s^{- 1}$ , is

A $0.24$
B $0.16$
C $0.30$
D $0.36$

Answer: D

Divide each rate by the coefficient:

\frac{1}{4} (- d [NH X_{3}] / d t) = \frac{1}{6} (d [H X_{2} O] / d t)

. Thus

d [H X_{2} O] / d t = (6/4) (0.24) = 0.36 M s^{- 1}

(

0.30

would be the rate of

O X_{2}

consumption, not of water formation).

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