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

Step-by-step explanation:

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

Also at 1500 K,
`K_(r)` =
`2.82 * 10^(5) * e^({(-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) * 8.314 * 1500`

`K_(c) = 2.9 * 10^(-7)`

Also,
`K_(c) = (K_(f))/(K_(r))`

or,
`K_(f) = K_(c) * K_(r)`

=
`2.9 * 10^(-7) * 400.613`

=
`1.1617 * 10^(-4) m^(6)/mol^(2)s`

Thus, we can conclude that the value of
`K_(f)` is
`1.1617 * 10^(-4) m^(6)/mol^(2)s`.