Question

for the reaction A + B reversible C if kp is 10 at 0°C and kf 2.24 × 10^-2 then find kb​

Answer 1

Answer:

The rate constant for the given backward reaction is $1.009\ *\ 10^{-4}$·

Explanation:

The reaction is given as,

    A  $+$  B   →   C

$K_p\ \ =\ \ 10$

Temperature, T      $=$     $0^{0}C\ \ =\ \ 273$ K

The forward rate constant, $K_f$    $=$     $2.24\ *\ 10^{-2}$

To calculate the rate constant for the backward reaction, $K_b$  first, we need to find out the value for the $K_c$ ·

For that, we have the relationship that

    $K_P\ =\ K_c \ \ (R\ T)^{\Delta n}$  $----(1)$

where,

$K_c$  and $K_p$ are the equilibrium constants·

R   $=$ The universal gas constant $=\ 0.0821\ L\ atm\ mol^{-1}\ K^{-1}$

$\Delta n \ =$ The No of moles of the product $-$ The No of moles of the reactant

Here,

$\Delta n \ =$  $1\ -\ 2\ =\ -1$ mol

On rearranging the equation No· $1$ with the term $K_c$ on the left-hand side,

we get,

    $K_c\ =\ \ \frac{K_p}{ (R\ T)^{\Delta n}}$

On substituting the values we get,

⇒ $K_c\ =\ \ \frac{10}{ (0.0821\ *\ 273)^{-1}}$

⇒ $K_c\ =\ \ \frac{10}{ (22.4133)^{-1}}$

⇒ $K_c\ =\ \ \frac{10}{ 0.045}$

⇒ $K_c\ =\ \ 2.22\ *\ 10^{2}$

Now, to calculate the $K_b$ we have the relationship that

   $\frac{K_f}{K_b}\ =\ K_c$

On rearranging we get,

   $K_b\ =\ \frac{K_f}{K_c}$

On substituting the values we get,

⇒ $K_b\ =\ \frac{2.24\ *\ 10^{-2} }{2.22\ *\ 10^{2} }$

⇒ $K_b\ =\ 1.009\ *\ 10^{-4}$

Therefore,

The rate constant for the given backward reaction is $1.009\ *\ 10^{-4}$·