Question

In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691 based on a reference-state pressure of 1 atm (101,325 Pa). Derive an algebraic ex- pression for the forward rate coefficient kf . Evaluate your expression for a temperature of 1500 K. Give units.

Answer 1

Explanation:

Let us assume that the value of $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s$

Also at 1500 K, $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s$

                     $K_{r} = 400.613 m^{6}/mol^{2}s$

Relation between $K_{p}$ and $K_{c}$ is as follows.

                  $K_{p} = K_{c}RT$

Putting the given values into the above formula as follows.

                  $K_{p} = K_{c}RT$

         $0.003691 = K_{c} \times 8.314 \times 1500$

                $K_{c} = 2.9 \times 10^{-7}$

Also,     $K_{c} = \frac{K_{f}}{K_{r}}$

or,                $K_{f} = K_{c} \times K_{r}$

                               = $2.9 \times 10^{-7} \times 400.613$

                               = $1.1617 \times 10^{-4} m^{6}/mol^{2}s$

Thus, we can conclude that the value of $K_{f}$ is $1.1617 \times 10^{-4} m^{6}/mol^{2}s$.