Question

The decomposition of a hydrocarbon follows the equation ${\mathbf{k}} = \left( {{\mathbf{4}}.{\mathbf{5}} \times {\mathbf{1}}{{\mathbf{0}}^{{\mathbf{11}}}}{{\mathbf{s}}^{ - {\mathbf{1}}}}} \right){\mathbf{e}}{ - ^{{\mathbf{28000}}/{\mathbf{K}}/{\mathbf{T}}}}$. What will be the value of activation energy?
A) ${\mathbf{669}}{\text{ kJ}}{\mathbf{Mo}}{{\mathbf{l}}^{ - {\mathbf{1}}}}$
B) ${\mathbf{232}}.{\mathbf{79}}{\text{ }}{\mathbf{kJMo}}{{\mathbf{l}}^{ - {\mathbf{1}}}}$
C) ${\mathbf{4}}.{\mathbf{5}} \times {\mathbf{1}}{{\mathbf{0}}^{{\mathbf{11}}}}{\mathbf{kJMo}}{{\mathbf{l}}^{ - {\mathbf{1}}}}$
D) ${\mathbf{28000}}{\text{ }}{\mathbf{kJMo}}{{\mathbf{l}}^{ - {\mathbf{1}}}}$

Answer 1

In the given question we have to calculate the value of activation energy of the decomposed hydrocarbon, by the given equation and the value of rate constant. By taking help of Arrhenius' equation one can easily calculate the value of activation energy.

Formula used: $k = {\text{ }}A{e^{ - Ea}}^{/RT\;\;}$
Where $A$= the pre-exponential factor for the reaction
$R$= the universal gas constant,
$T$= the absolute temperature,
$k$= the reaction rate coefficient
${E_a}$= activation energy

Complete step-by-step answer:
The decomposition of a hydrocarbon might help to reduce greenhouse gases by means of co-producing valuable carbon products such as carbon black and graphite-like carbon products.
Activation energy, in terms of chemistry, can be defined as the minimum amount of energy that is required to activate atoms or molecules to a position in which they can experience chemical transformation.
From the given question the decomposition of hydrocarbon occur by the following equation
$k = {\text{ }}\left( {4.5 \times {{10}^{11}}{s^{ - 1}}} \right){\text{ }}{e^{- 2800\dfrac{k}{T}}}$
So, we have to calculate activation energy ${E_a}$
The given equation is
$k = {\text{ }}\left( {4.5 \times {{10}^{11}}{s^{ - 1}}} \right){\text{ }}{e^{- 2800\dfrac{k}{T}}}$ (1)
According to Arrhenius equation we get;
$k = {\text{ }}A{e^{\; - {E_a}}}^{/RT\;\;}$ (2)
From equation (1) and (2), we get; $$$$
$\dfrac{{{E_a}}}{{RT\;}}{\text{ }} = \;{\text{ }}\dfrac{{28000K}}{T}$
${E_a} = {\text{ }}R \times 28000K\;$
$= {\text{ }}8.314{\text{ }}J{\text{ }}{K^{\; - {\text{ }}1}}mo{l^{\; - {\text{ }}1}} \times {\text{ }}28000{\text{ }}K$
$= {\text{ }}232792{\text{ k}}J{\text{ }}mol{\;^{ - {\text{ }}1}}$
$$$$$$= {\text{ }}232.792{\text{ }}kJ{\text{ }}mo{l^{; - {\text{ }}1}}$$

Hence, the correct answer is option ‘B’.

Note: Decomposition of hydrocarbons to Carbon dioxide -free hydrogen and elemental carbon takes place through the following process which is first step is single-step catalysis, second is thermal cracking, or can be plasma decomposition or decarbonization which follows splitting and dissociation of hydrocarbons. As we know that the Arrhenius equation gives the quantitative basis of the relationship between the activation energy and the rate at which a reaction occurs.