Consider the reaction, $N_{2} (g) + 3 H_{2} (g) ⟶ 2 N H_{3} (g)$ . The equality relationship between $\frac{d [N H_{3}]}{d t}$ and $- \frac{d [H_{2}]}{d t}$ is:

\frac{d [N H_{3}]}{d t} = - \frac{1}{3} \frac{d [H_{2}]}{d t}

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+ \frac{d [N H_{3}]}{d t} = - \frac{2}{3} \frac{d [H_{2}]}{d t}

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+ \frac{d [N H_{3}]}{d t} = - \frac{3}{2} \frac{d [H_{2}]}{d t}

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\frac{d [N H_{3}]}{d t} = - \frac{d [H_{2}]}{d t}

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Solution

The correct option is A $+ \frac{d [N H_{3}]}{d t} = - \frac{2}{3} \frac{d [H_{2}]}{d t}$
$- \frac{1}{3} \frac{d [H_{2}]}{d t} = \frac{1}{2} \frac{d [{N H}_{3}]}{d t} \frac{d [{N H}_{3}]}{d t} = - \frac{2}{3} \frac{d [H_{2}]}{d t}$

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Similar questions

Q. For the chemical reaction

N_{2}

(g) +

3 H_{2}

(g)

⇌

2 N H_{3}

(g)
The correct option is:

Consider the chemical reaction,
$N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$
The rate of this reaction can be expressed in terms of time derivatives of concentration of $N_{2} (g), H_{2} (g) o r N H_{3} (g)$ . Identify the correct relationship amongst the rate expressions.

Consider the chemical reaction

$N_{2} (g) + 3 H_{2} (g) ⟶ 2 N H_{3} (g) .$

The rate of this reaction can be expressed in terms of time derivatives of the concentration of $N_{2} (g), H_{2} (g), o r N H_{3} (g) .$ Identify the correct relationship among the rate expressions.

Q. For the reaction

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

- \frac{d [N H_{3}]}{d t} = k_{1} [N H_{3}], \frac{d [N_{2}]}{d t} = k_{2} [N H_{3}], \frac{d [H_{2}]}{d t} = k_{3} [N H_{3}]

then the relation between

k_{1}, k_{2}

and

k_{3}

is?

Q. For a reaction,

2 N H_{3} ⟶ N_{2} + 3 H_{2}

, it is observed that:

\frac{- d [N H_{3}]}{d t} = k_{1} [N H_{3}], \frac{d [N_{2}]}{d t} = k_{2} [N H_{3}], \frac{d [H_{2}]}{d t} = k_{3} [N H_{3}]

What is the relation between

k_{1}, k_{2}

and

k_{3}

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Consider the reaction, N2(g)+3H2(g)⟶2NH3(g). The equality relationship between d[NH3]dt and −d[H2]dt is:

The correct option is A +d[NH3]dt=−23d[H2]dt−13d[H2]dt=12d[NH3]dtd[NH3]dt=−23d[H2]dt

Consider the reaction, $N_{2} (g) + 3 H_{2} (g) ⟶ 2 N H_{3} (g)$ . The equality relationship between $\frac{d [N H_{3}]}{d t}$ and $- \frac{d [H_{2}]}{d t}$ is:

The correct option is A $+ \frac{d [N H_{3}]}{d t} = - \frac{2}{3} \frac{d [H_{2}]}{d t}$
$- \frac{1}{3} \frac{d [H_{2}]}{d t} = \frac{1}{2} \frac{d [{N H}_{3}]}{d t} \frac{d [{N H}_{3}]}{d t} = - \frac{2}{3} \frac{d [H_{2}]}{d t}$