# Rate constant at temperature

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.

• Chemistry: calculating the rate constant for a reaction? the initial method was used to investigate the reaction: 2H2 + 2NO --> 2H2O + N2 – stephanie Feb 22 at 23:23
• 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? – orthocresol Feb 23 at 0:57
• The numbers given must be at 973 K as no other information is given ? – porphyrin Feb 23 at 14:42
• @porphyrin the information is the table – stephanie Feb 24 at 0:54
• There is no mention of different temperatures in the table. All the table gives is $k=0.384$ with the rate equation you use.. – porphyrin Feb 24 at 8:14