Discussion Forum

At 518° C,  the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr, was 1.00 Torr s^-1 when 5% had reacted and 0.5 Torr s^-1  when 33% had reacted. The order of the reaction is

Answer:

Explanation:

For the reaction

$CH_{3}CHO(g)$        $\underrightarrow{Decomposes}$                 $CH_{4}+CO$

Let the order of reaction with respect to CH₃CHO is m.

   Its given, r₁=1 torr/sec. when CH₃CHO is 5% reacted i.e 95% unreacted . Similarly r₂= 0.5 torr/sec when CH₃CHO  is 33% reacted i.e. , 67% unreacted.

    Use the formula,   $r\alpha(a-x)^{m}$

where $(a-x)$ =amount unreacted

so,      $\frac{r_{1}}{r_{2}}=\frac{(a-x_{1})^{m}}{(a-x_{2})^{m}}$ or 

                               $\frac{r_{1}}{r_{2}}=\left[\frac{(a-x_{1})^{}}{(a-x_{2})^{}}\right]^{m}$

Now putting the given values

      $\frac{1}{0.5}=( \frac{0.95}{0.67})^{m}$

$\Rightarrow$        $2= (1.41)^{m}$           or           m=  2