From the following data find a satisfactory rate equation for the gas-phase decomposition of pure A, A -> R + S, in a mixed flow reactor.

τ based on inlet feed conditions, sec | 0.423 | 5.1 | 13.5 | 44.0 | 192 |

X_{A} (for C_{Ao} = 0.002 mol/lit) | 0.22 | 0.63 | 0.75 | 0.88 | 0.96 |

Calculations:

For the mixed flow reactor, the design equation is,

τ = C_{Ao}X_{A}/-r_{A} → 1

For variable density systems,

Where, ε_{A} = fractional change in volume = (V_{XA = 1} - V_{XA = 0}) / V_{XA = 0} = (2 - 1)/1 = 1

Therefore,

à 2

* For the first order equation*, equn.1 can be written as

τ = C_{Ao}X_{A}/ kC_{A} → 3

substituting for C_{Ao} in equation 3,

τ = 0.002 X_{A}/ kC_{A}

substituting for C_{A} from equn.2,

That is, if the given data is for a first order reaction, τ vs. X_{A}(1 + X_{A})/(1 - X_{A}) will be a straight line, with a slope of k.

τ, sec |
0.423 |
5.1 |
13.5 |
44 |
192 |

X |
0.22 |
0.63 |
0.75 |
0.88 |
0.96 |

X |
0.3441 |
2.7754 |
5.25 |
13.7867 |
47.04 |

From the above data, the following graph is drawn :(graph 1)

** For the second order equation**, equn.1 can be written as

τ = C_{Ao}X_{A}/ kC_{A}^{2}

That is, if the given data is for a second order reaction, τ vs. X_{A}(1 + X_{A})^{2}/ (1 - X_{A})^{2} will be a straight line with a slope of kC_{Ao}.

τ, sec |
0.423 |
5.1 |
13.5 |
44 |
192 |

X |
0.22 |
0.63 |
0.75 |
0.88 |
0.96 |

X |
0.5382 |
12.2268 |
36.75 |
215.99 |
2304.96 |

From the above data, the following graph is drawn :(graph 2)

By comparing two graphs it is seen that first order kinetics is well fitting the given data.

That is, the reaction is following first order.

Slope = 0.2493 (from the graph.1)

Therefore,

k = 0.2493 sec^{-1}.

Therefore, rate equation for the given reaction is

**-r _{A}** = 0.2493 C

Last Modified on: 30-Apr-2024

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