In a reaction A_2B_3(g) to A_2(g) + frac{3}{2 | vedclass

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Question

$\begin{array}{*{20}{c}}
  {}&{\,\,\,{A_2}{B_{3\,}}\,\,\,\,\, 	o }&{{A_2}\,\,\,\, + }&{\frac{3}{2}\,{B_2}} \ 
  {t\, =

Vedclass · Accepted Answer

$4$

Vedclass · Answer

$\begin{array}{*{20}{c}}
  {}&{\,\,\,{A_2}{B_{3\,}}\,\,\,\,\, 	o }&{{A_2}\,\,\,\, + }&{\frac{3}{2}\,{B_2}} \ 
  {t\, = \,0}&{{P_0}\,\,\,\,\,}&{0\,\,\,\,}&0 \ 
  {eq.}&{{P_0} - x\,\,\,\,}&{x\,\,\,\,}&{\frac{{3x}}{2}} 
\end{array}$

$75=P_{0}+\frac{3}{2} x$      $ \Rightarrow  \frac{3 x}{2}=75-60$

$x=10$

$	ext { Rate } =\left|\frac{\Delta p}{\Delta t}ight|=\left|\frac{50-60}{2.5}ight|$

$=4 \,	ext { torr } \,\min ^{-1}$

	$1$	$2$	$3$
$\mathrm{HI}\left(\mathrm{mol} \mathrm{L}^{-1}\right)$	$0.005$	$0.01$	$0.02$
Rate $\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}-1\right)$	$7.5 \times 10^{-4}$	$3.0 \times 10^{-3}$	$1.2 \times 10^{-2}$

$p ( mm Hg )$	$50$	$100$	$200$	$400$
Relative $t _{1 / 2}( s )$	$4$	$2$	$1$	$0.5$

In a reaction $A_2B_3(g) \to A_2(g) + \frac{3}{2}B_2(g)$, the pressure increases from $60$ torr to $75$ torr in $2.5\, minutes$. The rate of disappearance of $A_2B_3$ is ........ $torr\, min^{-1}$

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