Question

For a gas reaction at $T ( K )$ the rate is given $by -\frac{d p_{A}}{d t}$ $=k' p_{A}^{2}$ atm/hr. If the rate equation is expressed as :
$- r _{ A }=-\frac{1}{V} \frac{d n_{A}}{d t}=k C_{A}^{2}, mol /($ litre-hr $)$, the rate constant $k$ is given by-
where $R =$ ideal gas law constant, cal/g mol. $K$

• A. $k = k '$
• B. $k = k ' RT$
• C. $k = k ' / RT$
• D. $k = k '( RT )^{2}$

Answer 1

Answer: (D)

Given, $ r=-\frac{d P_{A}}{d t}=k'P_{A}^{2} \,atm / \,hr$

$\&\,\,\, r =-\frac{1}{V} \frac{d n_{A}}{d t}=k C_{A}^{2} \,Mhr ^{-1}$

$\therefore k '\,\, P_{A}^{2}=k C_{A}^{2}$

$\Rightarrow k = k '\left(\frac{P_{A}}{C_{A}}\right)^{2}= k '( R T )^{2}$