For a gas reaction at T ( K ) the rate is given by -(d pA/d t) =k' pA2 atm/hr. If the rate equation is expressed as: - r A =-(1/V) (d nA/d t)=k CA2, mol /( litre-hr ), the rate constant k is given by- where R = ideal gas law constant, cal/g mol. K

Question

For a gas reaction at T ( K ) the rate is given by -(d pA/d t) =k' pA2 atm/hr. If the rate equation is expressed as: - r A =-(1/V) (d nA/d t)=k CA2, 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

Tardigrade Admin · Accepted Answer

Given, r=-(d PA/d t)=k'PA2 atm / hr r =-(1/V) (d nA/d t)=k CA2 Mhr -1 ∴ k ' PA2=k CA2 ⇒ k = k '((PA/CA))2= k '( R T )2

$k = k^{'}$

$k = k^{'} RT$

$k = k^{'} / RT$

$k = k^{'} (RT)^{2}$

Given, $r = - \frac{d P _{A}}{d t} = k^{'} P_{A}^{2} a t m / h r$

$& r = - \frac{1}{V} \frac{d n _{A}}{d t} = k C_{A}^{2} M h r^{- 1}$

$∴ k^{'} P_{A}^{2} = k C_{A}^{2}$

$\Rightarrow k = k^{'} (\frac{P _{A}}{C _{A}})^{2} = k^{'} (RT)^{2}$

Q. For a gas reaction at T(K) the rate is given by−dtdpA​​ =k′pA2​ atm/hr. If the rate equation is expressed as : −rA​=−V1​dtdnA​​=kCA2​,mol/( litre-hr ), the rate constant k is given by- where R= ideal gas law constant, cal/g mol. K

k=k′

k=k′RT

k=k′/RT

k=k′(RT)2

Given, r=−dtdPA​​=k′PA2​atm/hr &r=−V1​dtdnA​​=kCA2​Mhr−1 ∴k′PA2​=kCA2​ ⇒k=k′(CA​PA​​)2=k′(RT)2

$k = k^{'}$

$k = k^{'} RT$

$k = k^{'} / RT$

$k = k^{'} (RT)^{2}$

Given, $r = - \frac{d P _{A}}{d t} = k^{'} P_{A}^{2} a t m / h r$

$& r = - \frac{1}{V} \frac{d n _{A}}{d t} = k C_{A}^{2} M h r^{- 1}$

$∴ k^{'} P_{A}^{2} = k C_{A}^{2}$

$\Rightarrow k = k^{'} (\frac{P _{A}}{C _{A}})^{2} = k^{'} (RT)^{2}$