JEE CHEMISTRY MCQ: CHEMICAL KINETICS

1 Easy

1 of 1

Understanding +4 | -1

Consider the reaction

{N_2}(g) + 3{H_2}(g) \to 2N{H_3}(g)

. The equality relationship between

\frac{{d\left[ {N{H_3}} \right]}}{{dt}}\alpha - \frac{{d\left[ {{H_2}} \right]}}{{dt}}

View answer and explanation

Correct option: A

+ \frac{{d\left[ {N{H_3}} \right]}}{{dt}} = \frac{{ - 2}}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}}

Option A is correct because the rate of formation of

NH_3

is related to the rate of consumption of

H_2

by the stoichiometric coefficients in the balanced chemical equation.

More Questions from CHEMICAL KINETICS

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\mathrm{A}_2+\mathrm{B}_2 \rightarrow 2 \mathrm{AB}

\begin{aligned} & {{\text{A}}_2}\mathop \rightleftharpoons \limits_{{{\text{k}}_{ - 1}}}^{{{\text{k}}_1}} {\text{A}} + {\text{A (fast) }} \\ & \mathrm{A}+\mathrm{B}_2 \xrightarrow{\mathrm{k}_2} \mathrm{AB}+\mathrm{B} \text { (slow) } \\ & \mathrm{A}+\mathrm{B} \rightarrow \mathrm{AB} \text { (fast) } \end{aligned}

k

(Initial\mathrm{ }conc.\mathrm{ } = a)

C_0 \sqrt{T}

C_0

T

100 \mathrm{~kJ/mol}.

75 \%

\log _{10} \frac{K_2}{K_1}

27^{\circ} \mathrm{C}

2.303 \times 8.314=19.147