Given
Mass of NO - 824 g
Molar mass of NO - 30.01g/mol
No of moles of NO = Given mass/Molar mass
No of moles of NO = 824/30.01= 27.45 mole
Hence 27.5 moles of NO are formed!
Answer:
(a)
(b)
Explanation:
Hello,
(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:
![\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BNOBr%5D%7D%3Dkt%20%2B%5Cfrac%7B1%7D%7B%5BNOBr%5D_0%7D%5C%5C%5Cfrac%7B1%7D%7B%5BNOBr%5D%7D%3D%5Cfrac%7B0.8%7D%7BM%2As%7D%2A22s%2B%5Cfrac%7B1%7D%7B0.086M%7D%3D%5Cfrac%7B29.3%7D%7BM%7D%5C%5C)
![[NOBr]=\frac{1}{29.2/M}=0.0342M](https://tex.z-dn.net/?f=%5BNOBr%5D%3D%5Cfrac%7B1%7D%7B29.2%2FM%7D%3D0.0342M)
(b) Now, for a second-order reaction, the half-life is computed as shown below:
![t_{1/2}=\frac{1}{k[NOBr]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5BNOBr%5D_0%7D)
Therefore, for the given initial concentrations one obtains:
Best regards.
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