JEE CHEMISTRY MCQ: CHEMICAL KINETICS

1 Hard

1 of 1

Analyzing +4 | -1

For the reaction

2 \mathrm{NH}_3 \rightarrow \mathrm{N}_2+3 \mathrm{H}_2

-\frac{\mathrm{d}\left[\mathrm{NH}_3\right]}{\mathrm{d} t}=k_1\left[\mathrm{NH}_3\right],-\frac{\mathrm{d}\left[\mathrm{N}_2\right]}{\mathrm{d} t}=k_2\left[\mathrm{NH}_3\right], \frac{\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{d} t}=k_3\left[\mathrm{NH}_3\right]

Then relation between

k_1, k_2

and

k_3

View answer and explanation

Correct option: A

1.5 k_1=3 k_2=k_3

Option A (

1.5 k_1=3 k_2=k_3

) is correct because it correctly relates the rate constants based on the stoichiometry of the reaction.

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