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3 years ago

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Which letter (a–f) represents potential energy (ΔH) of the products?

1 answer:

Eddi Din [679]3 years ago

8 0

Answer:

<em>1) Which letter (a–f) represents potential energy (ΔH) of the products?</em>

(e)

<em>2) Which letter (a-f) represents potential energy (ΔH) of the activated complex?</em>

(c)

<em>3) Which letter (a-f) represents potential energy (ΔH) of the reactants?</em>

(a)

Explanation:

The graph ploted represents the changes in the chemical potential (stored) energy for the reaction X + Y → Z as long as it evolves from reactants to products.

The left side of the graph shows the potential energy of the <em>reactants</em>, X and Y, when the reaction has not yet occurred.

Thus, the letter (a) is the potential energy of the reactants.

Moving from left to right you see that the potential energy increases, until a maximum. The maximum is when the reactants (X and y) have reacted up to an intermediate stage, in which there exists a transition state featured by an <em>activated complex</em> whose structure is half-way between the reactants and the products.

Thus, the letter (c) represents the potential energy of the activated complex.

At the far right of the graph, the reaction has gone to completion, the activated complex ends the transition state and the final products are formed. Thus, the letter (e) represents the potential energy of the products (Z).

All those energies are measured with respect to the same base line (the x-axis).

The attached graph has some labels added to help to identify the 3 stages: reactants, activated complex, and product.

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The reaction C 4 H 8 ( g ) ⟶ 2 C 2 H 4 ( g ) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ / mol. 262 kJ/mol. At 600.0 K,

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According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

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$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $600.0K$ = $6.1\times 10^{-8}s^{-1}$

$K_2$ = rate constant at $775.0$ = $?$

$Ea$ = activation energy for the reaction = 262 kJ/mol = 262000J/mol

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $600.0K$

$T_2$ = final temperature = $775.0K$

Now put all the given values in this formula, we get

$\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]$

$\log (\frac{6.1\times 10^{-8}s^}{K_2})=5.150$

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Therefore, the value of the rate constant at 775.0 K is $4.3\times 10^{-13}s^{-1}$

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