A substance X decomposes via two parallel pseudo-first-order pathways to Y and Z in a large excess of solvent W.
$X + W k 1 Y$ and $X + W k 2 Z$ .
The activation energies are $E_{a 1} = 50 kJ / mol$ and $E_{a 2} = 70 kJ / mol$ . At $300 K, k_{1} = 3 k_{2}$ . Which of the following is/are correct?

Question

A substance X decomposes via two parallel pseudo-first-order pathways to Y and Z in a large excess of solvent W.

$X + W k 1 Y$ and $X + W k 2 Z$ .

The activation energies are $E_{a 1} = 50 kJ / mol$ and $E_{a 2} = 70 kJ / mol$ . At $300 K, k_{1} = 3 k_{2}$ . Which of the following is/are correct?

TestHub · Accepted Answer

Concept:

Parallel reactions, Arrhenius equation, Effective activation energy $(E_{eff} = \frac{k _{1} E _{a 1} + k _{2} E _{a 2}}{k _{1} + k _{2}})$ , Yield analysis.

Solution:

For parallel reactions X → Y and X → Z, $k_{n e t} = k_{1} + k_{2}$ . Thus, (A) is correct.

$\frac{[ Y ]}{[ Z ]} = \frac{k _{1}}{k _{2}}$ . Since $k_{1}$ and $k_{2}$ are constants at a fixed temperature (T), this ratio is independent of time. Thus, (D) is correct.

A higher activation energy ( $E_{a}$ ) means the reaction is more sensitive to temperature. Since $E_{a 2} > E_{a 1}$ , $k_{2}$ increases more rapidly with T than $k_{1}$ . Thus, the yield of Z increases. Thus, (B) is correct.

Final Answer: Option (1, 2, 3 and 4)

$E_{e ff} = \frac{k _{1} E _{a_{1}} + k _{12} E a _{2}}{k _{1} + k _{2}}$ given $k_{1} = 3 k_{2}$

$E_{e ff} = \frac{3 k _{2} ( 50 ) + k _{2} ( 70 )}{4 k _{2}} = \frac{150 + 70}{4} = \frac{220}{4} = 55 k J m o l^{- 1}$

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Options:(select one or more)

Doubts & Discussion