The oxidation of ammonia happens by this reaction:
$4 N H_{3} (g) + 3 O_{2} (g) ⟶ 2 N_{2} (g) + 6 H_{2} O (g)$ , if the rate of formation of $N_{2}$ is $0.5 m o l L^{- 1} s^{- 1}$ then, the rate at which $N H_{3}$ gets consumed is:

1.0 m o l L^{- 1} s^{- 1}

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0.5 m o l L^{- 1} s^{- 1}

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2.0 m o l L^{- 1} s^{- 1}

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1.5 m o l L^{- 1} s^{- 1}

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Solution

The correct option is A $1.0 m o l L^{- 1} s^{- 1}$
The rate of consumption of $N H_{3} = - \frac{d [N H_{3}]}{d t}$
The rate of formation of $N_{2} = + \frac{d [N_{2}]}{d t}$ = $0.5 m o l L^{- 1} s^{- 1}$
The rate of the reaction can be written as:
$\frac{- 1}{4} \frac{d [N H_{3}]}{d t} = - \frac{1}{3} \frac{d [O_{2}]}{d t} = \frac{1}{2} \frac{d [N_{2}]}{d t} = \frac{1}{6} \frac{d [H_{2} O]}{d t}$
$\frac{- 1}{4} \frac{d [N H_{3}]}{d t} = \frac{1}{2} \frac{d [N_{2}]}{d t}$
$\frac{- d [N H_{3}]}{d t} = 4 \times \frac{1}{2} \times 0.5 = 1.0 m o l L^{- 1} s^{- 1}$

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Similar questions

Q. The oxidation of ammonia happens by this reaction:

4 N H_{3} (g) + 3 O_{2} (g) ⟶ 2 N_{2} (g) + 6 H_{2} O (g)

, if the rate of formation of

N_{2}

0.5 m o l L^{- 1} s^{- 1}

then, the rate at which

N H_{3}

gets consumed is:

Q. For the reaction,

2 S O_{2} + O_{2} ⇌ 2 S O_{3}

, the rate of disappearance of

O_{2}

2 \times 10^{- 4} m o l L^{- 1} s^{- 1}

. The rate of appearance of

S O_{3}

is:

For the reaction, $N_{2} O_{5} (g) \to N O_{2} (g) + \frac{1}{2} O_{2} (g)$ The value of rate of disappearance of $N_{2} O_{5}$ is given as $6.25 \times 10^{- 3} m o l L^{- 1} s^{- 1} .$ The rate of formation of $N O_{2}$ and $O_{2}$ is given respectively as:

Q. The rate of a first order reaction

R \to P

7.5 \times 10^{- 4} m o l L^{- 1} s^{- 1}

when the concentration of R is 0.5 mol

L^{- 1}

.

The rate constant is

Q. The plot of

t_{1 / 2} v / s [R]_{0}

for a reaction is a straight-line parallel to x-axis. The unit for the rate constant of this reaction is:

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The oxidation of ammonia happens by this reaction: 4NH3(g)+3O2(g)⟶2N2(g)+6H2O(g), if the rate of formation of N2 is 0.5 molL−1s−1 then, the rate at which NH3 gets consumed is:

The correct option is A 1.0 molL−1s−1The rate of consumption of NH3=−d[NH3]dt The rate of formation of N2=+d[N2]dt = 0.5 molL−1s−1 The rate of the reaction can be written as: −14d[NH3]dt=−13d[O2]dt=12d[N2]dt=16d[H2O]dt −14d[NH3]dt=12d[N2]dt −d[NH3]dt=4×12×0.5=1.0 molL−1s−1

The oxidation of ammonia happens by this reaction:
$4 N H_{3} (g) + 3 O_{2} (g) ⟶ 2 N_{2} (g) + 6 H_{2} O (g)$ , if the rate of formation of $N_{2}$ is $0.5 m o l L^{- 1} s^{- 1}$ then, the rate at which $N H_{3}$ gets consumed is: