Question

In the following reaction; x → AyB \({\rm{lo}}{{\rm{g}}_{10}}\left[ {\frac{{ - d\left[ A \right]}}{{dt}}{\rm{\;}}} \right] = {\rm{lo}}{{\rm{g}}_{10}}\left[ {\frac{{d\left[ B \right]}}{{dt}}} \right] + 0.3010\)

‘A’ and ‘B’ respectively can be:
1. n-Butane and Iso-butane
2. C₂H₂ and C₆H₆
3. C₂H₄ and C₄H₈
4. N₂O₄ and NO₂

Answer 1

Correct Answer - Option 3 : C₂H₄ and C₄H₈

Concept:

Given reaction xA → yB

\(\frac{{ - dA}}{{xdt}} = \frac{1}{y}\frac{{{\rm{dB}}}}{{dt}}\) 

\(\frac{{ - d\left[ A \right]}}{{dt}} = \frac{{{\rm{dB}}}}{{dt}} \times \frac{x}{y}\) 

\({\rm{log}}\frac{{ - dA}}{{dt}} = {\rm{log}}\left[ {\frac{{dB}}{{dt}}} \right] + {\rm{log\;}}\left( {\frac{x}{y}} \right)\) 

Where negative sign indicates rate of disappearance of the reactant.

On comparing this equation with the given equation in question. We get,

\({\rm{log\;}}\left( {\frac{x}{y}} \right) = 0.3010\;{\rm{or}}\;{\rm{log}}\left( {\frac{x}{y}} \right) = {\rm{log}}\;2\) 

\(\left( {\frac{x}{y}} \right) = 2\) 

∴ The reaction is a type 2A → B

2C₂H₄ (Ethylene) → C₄H₈ (1-Butene)

Ethylene 2C₂H₄ is an important industrial organic chemical. It is produced by heating either natural gas, especially its ethane and propane components, or petroleum to 800-900°C, giving a mixture of gases from which, the ethylene is separated.

1-Butene C₄H₈ is an organic chemical compound, linear alpha-olefin (alkene) and one of the isomers of butene. The formula is CH₃CH₂CH=CH₂. It is a highly flammable, easily condensed gas.

1-Butene is stable in itself but polymerizes readily to polybutene. Its main application is as a co-monomer in the production of certain kinds of polyethylene, such as linear low-density polyethylene (LLDPE). It has also been used as a precursor to polypropylene resins, butylene oxide and butanone.

Hence, option (c) is correct.