Question

For a gaseous reaction at T( k),the rate is given by -d(PA)/dt=k'PA^2 atm/hr.If rate equation is expressed as -rA=(-1/V)d(nA)/dt=kCA^2 mol/l/hr, then relationship between K and k'is

Answer 1

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Answer 2

Given:

For a gaseous reaction, $-\frac{dp_A}{dt} = K'p_A^2$

Rate equation, $-r_A = -\frac{1}{V}\frac{dn_A}{dt} = Kc_A^2$

To find:

Relationship between K and K'.

Solution:

We have given,

          $-\frac{dP_A}{dt} = K'P_A^2$             -(1)

and

            $-r_A = -\frac{1}{V}\frac{dn_A}{dt} = Kc_A^2$        -(2)

For an ideal gas,

                $P_AV = n_ART$

therefore,

                $n_A = \frac{P_AV}{RT}$

by putting above value of $n_A$ in equation (2), we get

      $-r_A = -\frac{1}{V}\frac{d}{dt}\frac{P_AV}{RT}$

now, substitute the value of from eq.(1) in above equation, we get,

     $-r_A = -\frac{1}{V}\frac{K'P_A^2V}{RT} = Kc_A^2$       -(3)

We know that, $P_A = \frac{n_A}{V}RT$

so, $P_A = c_ART$

squaring both sides,

   $c_A^2= \frac{P_A^2}{R^2T^2}$

now, putting above value in equation (3),

  $-r_A = -\frac{1}{V}\frac{K'P_A^2V}{RT} = K\frac{P_A^2}{R^2T^2}$

By solving this, we get

  $K = K' RT$

which is a required relation.