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So, I don't know where I have went wrong.

The question is:

The initial method was used to investigate the reaction: $$\ce{2H2 + 2NO -> 2H2O + N2}$$

$$ \begin{array}{clrr} \hline & \ce{H2} & \pu{10^-2} & \pu{mol dm^-3} & 2.0 & 2.0 & 2.0 & 1.0 & 4.0 \\ \hline & \ce{NO} & \pu{10^-2} & \pu{mol dm^-3} & 2.50 & 1.25 & 5.00 & 1.25 & 2.50 \\ \hline & rate & \pu{10^-6} & \pu{mol dm^-3 s^-1} & 4.8 & 1.2 & 19.2 & 0.6 & 9.6 \\ \hline \end{array} $$

$$ rate = k \ce{[H2][NO]^2 }$$

Calculate the value for the rate constant for this reaction at $\pu{973 K}$.

My working out: I used the equation $$ rate=\pu{k[A][B]^2} $$

$$ \pu{4.8 \times 10^{-6}} = k \pu{( 2.0 \times 10^{-2})( 2.50 \times 10^{-2})^{2}} $$

so I rearranged and got this:

$$ k= \pu{ {4.8 \times 10^{-6}}\over {{( 2.0 \times 10^{-2}) ( 2.50 \times 10^{-2})^{2}}} = 0.384}$$

then i did: $$\pu{ {0.384}\over {973} = 3.94655704 \times 10^{-4}}$$

Can someone please tell me where I've gone wrong? I'd be much appreciated.

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  • $\begingroup$ Chemistry: calculating the rate constant for a reaction? the initial method was used to investigate the reaction: 2H2 + 2NO --> 2H2O + N2 $\endgroup$ – stephanie Feb 22 at 23:23
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    $\begingroup$ Your text is mostly indecipherable, because of the formatting; but my first suggestion would be to include units in your working. It might seem like a chore, but it is important. Please see How do I format my posts using Markdown or HTML? and FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange? $\endgroup$ – orthocresol♦ Feb 23 at 0:57
  • $\begingroup$ The numbers given must be at 973 K as no other information is given ? $\endgroup$ – porphyrin Feb 23 at 14:42
  • $\begingroup$ @porphyrin the information is the table $\endgroup$ – stephanie Feb 24 at 0:54
  • $\begingroup$ There is no mention of different temperatures in the table. All the table gives is $k=0.384$ with the rate equation you use.. $\endgroup$ – porphyrin Feb 24 at 8:14

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