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The rate of the reaction: CH(3)COOC(2)...

The rate of the reaction:
`CH_(3)COOC_(2)H_(5) + NaOH to Ch_(3)COONa + C_(2)H_(5) OH` is given by the equation,
rate `= k [CH_(3) COOC_(2)H_(5)][NaOH]`
If concenration is expressed in mol/L the units of k are

A

`mol^(-2) L^(2) s^(-1)`

B

`mol L^(-1) s^(-1)`

C

`L mol^(-1) s^(-1)`

D

`S^(-1)`

Text Solution

AI Generated Solution

The correct Answer is:
To determine the units of the rate constant \( k \) for the given reaction, we start with the rate equation provided: \[ \text{rate} = k [CH_3COOC_2H_5][NaOH] \] ### Step 1: Understand the Units of Rate The rate of a reaction is typically expressed in terms of concentration change over time. In this case, the rate can be expressed in units of molarity per second (mol/L/s). Therefore, the units of rate are: \[ \text{Units of Rate} = \frac{\text{mol}}{\text{L} \cdot \text{s}} = \text{mol} \cdot \text{L}^{-1} \cdot \text{s}^{-1} \] ### Step 2: Identify the Concentration Terms The concentrations of the reactants \( [CH_3COOC_2H_5] \) and \( [NaOH] \) are both expressed in molarity, which has units of mol/L. Thus, we can write: \[ [CH_3COOC_2H_5] = \frac{\text{mol}}{\text{L}} \quad \text{and} \quad [NaOH] = \frac{\text{mol}}{\text{L}} \] ### Step 3: Substitute Concentration Units into the Rate Equation Substituting the concentration units into the rate equation gives: \[ \text{rate} = k \left(\frac{\text{mol}}{\text{L}}\right) \left(\frac{\text{mol}}{\text{L}}\right) = k \left(\frac{\text{mol}^2}{\text{L}^2}\right) \] ### Step 4: Rearrange to Solve for \( k \) Now, we can rearrange the equation to solve for \( k \): \[ k = \frac{\text{rate}}{[CH_3COOC_2H_5][NaOH]} = \frac{\text{mol} \cdot \text{L}^{-1} \cdot \text{s}^{-1}}{\frac{\text{mol}^2}{\text{L}^2}} \] ### Step 5: Simplify the Expression When we simplify the expression for \( k \): \[ k = \frac{\text{mol} \cdot \text{L}^{-1} \cdot \text{s}^{-1} \cdot \text{L}^2}{\text{mol}^2} = \frac{\text{L} \cdot \text{s}^{-1}}{\text{mol}} \] Thus, the units of \( k \) can be expressed as: \[ k = \frac{\text{L}}{\text{mol} \cdot \text{s}} = \text{L} \cdot \text{mol}^{-1} \cdot \text{s}^{-1} \] ### Conclusion The correct units for the rate constant \( k \) in this reaction are: \[ \text{L} \cdot \text{mol}^{-1} \cdot \text{s}^{-1} \] This corresponds to option 3: **liter per mole per second**. ---

To determine the units of the rate constant \( k \) for the given reaction, we start with the rate equation provided: \[ \text{rate} = k [CH_3COOC_2H_5][NaOH] \] ### Step 1: Understand the Units of Rate The rate of a reaction is typically expressed in terms of concentration change over time. In this case, the rate can be expressed in units of molarity per second (mol/L/s). Therefore, the units of rate are: ...
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The rate equation for the reaction CH_3COOC_2H_5+H_2O to^([H^+]) CH_3COOH+C_2H_5OH is: (dx)/(dt)=k[CH_3 COOC_2H_5][H_2O]^0 If the concentration of ester is doubled the rate becomes

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The equilibrium constant for the reaction: CH_(3)COOH(l) +C_(2)H_(5)OH(l) hArr CH_(3)COOC_(2)H_(5)(l)+H_(2)O(l) has been found to be equal to 4 at 25^(@)C . Calculate the free energy change for the reaction.

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