Question

If the percentage decompostion in $100$ minutes at the same temperature but at an initial pressure of $600$ mm is $(70+x)$ find the value of $x$__________.

Answer 1

$\because K=\frac{2.303}{t}log\frac{a}{(a-X)}$

Case I: $a\propto 200mm$;

$X\propto 200\times \left(50/100\right)mm$ and $t_{1/2}=53minute$

$\therefore K_1=\frac{2.303}{53}log\frac{200}{200-100}$

$=1.307\times {10}^{-2}minut{e}^{-1}$

Case II: $a\propto 200mm;X\propto 200\times (73\times 100)\propto 146mm$

$\therefore K_2=\frac{2.303}{100}log\frac{200}{200-146}$

$=1.309\times {10}^{-2}minut{e}^{-1}$

For a Ist order reaction, $t_{\left(1/n\right)}\propto (a{)}^0$ and therefore if initial pressure is 600 mm, the decomposition in 100 minutes will be 73%.