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  • Nitric oxide reacts with oxygen to form Nitrogen dioxide: <img alt="2 \mathrm{NO(g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})" src="https://entrancecorner.oncodecogs.c

Nitric oxide reacts with oxygen to form Nitrogen dioxide: 2 \mathrm{NO(g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})
Information about the reaction as mentioned below:
\begin{array}{ccc} {[\mathrm{NO}]_{\mathrm{mol}/L}} & {\left[\mathrm{O}_{2}\right]_{\mathrm{mol/L}}} & \text { Rate of Reaction mol} ^{-1}L\, min^{-1} \\ 0.04 & 0.04 & 4 \times 10^{-5} \\ 0.08 & 0.04 & 1.6 \times 10^{-4} \\ 0.04 & 0.08 & 8 \times 10^{-5} \end{array}

What is the value of rate constant for the reaction in proper unit?

Option: 1

0.5


Option: 2

0.625


Option: 3

5


Option: 4

625


Answers (1)

best_answer

Rate of Reaction \mathrm{=K\left[\mathrm{NO]}^{x}\left[\mathrm{O}_{2}\right]^{y}\right.}

By doubling \mathrm{[NO],} Rate quadruples, so \mathrm{x=2}
By doubling \mathrm{\left [ O_{2} \right ]}, Rate doubles , So \mathrm{y= 1}

Now, \mathrm{4 \times 10^{-5}=k(0.04)^{2}(0.04)^{1} \Rightarrow k=0.625}

Posted by

Gaurav

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