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The forward rate constant for the elemen...

The forward rate constant for the elementary reversible gaseous reaction
`C_(2)H_(6)implies2CH_(3)`
`"is"` `1.57xx10^(-3)s^(-1)at 100 K`
What is the rate constant for the backward reaction at this temperature if `10^(-4)` moles of `CH_(3)` and 10 moles of `C_(2)H_(6)` are present in a 10 litre vessel at equilibrium .

A

`1.57xx10^(9) L "mole"^(-1)s^(-1)`

B

`1.57xx10^(10) L "mole"^(-1)s^(-1)`

C

`1.57xx10^(11) L "mole"^(-1)s^(-1)`

D

`1.57xx10^(7) L "mole"^(-1)s^(-1)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the rate constant for the backward reaction of the given equilibrium reaction: \[ C_2H_6 \rightleftharpoons 2 CH_3 \] ### Step-by-Step Solution: 1. **Identify the Given Information**: - Forward rate constant (\( k_f \)) = \( 1.57 \times 10^{-3} \, s^{-1} \) - Moles of \( CH_3 \) at equilibrium = \( 10^{-4} \) moles - Moles of \( C_2H_6 \) at equilibrium = \( 10 \) moles - Volume of the vessel = \( 10 \, L \) 2. **Calculate the Concentrations**: - Concentration of \( CH_3 \): \[ [CH_3] = \frac{10^{-4} \, \text{moles}}{10 \, \text{L}} = 10^{-5} \, \text{M} \] - Concentration of \( C_2H_6 \): \[ [C_2H_6] = \frac{10 \, \text{moles}}{10 \, \text{L}} = 1 \, \text{M} \] 3. **Write the Expression for Equilibrium Constant (\( K_{eq} \))**: \[ K_{eq} = \frac{[CH_3]^2}{[C_2H_6]} \] 4. **Substitute the Concentrations into the \( K_{eq} \) Expression**: \[ K_{eq} = \frac{(10^{-5})^2}{1} = 10^{-10} \] 5. **Relate \( K_{eq} \) to the Rate Constants**: \[ K_{eq} = \frac{k_f}{k_b} \] Rearranging gives: \[ k_b = \frac{k_f}{K_{eq}} \] 6. **Substitute the Known Values**: \[ k_b = \frac{1.57 \times 10^{-3}}{10^{-10}} = 1.57 \times 10^{7} \, s^{-1} \] ### Final Answer: The rate constant for the backward reaction at 100 K is: \[ k_b = 1.57 \times 10^{7} \, s^{-1} \]
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