The rate law for the reaction: Ester+`H^(+)` Acid + Alcohol is
` (dx)/(dt) = k ` [Ester ]`[H^(+)]^(0)`
What would be the effect on the rate if
concentration of ester is doubled

Explanation

To determine the effect on the rate of the reaction when the concentration of ester is doubled, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Rate Law**: The rate law for the reaction is given as:
\[ \frac{dx}{dt} = k [\text{Ester}]^1 [H^+]^0 \]
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