Relating the different ways of expressing reaction
rates: Ethanol `(CH_(3)CH_(2)OH)`, the avtive ingredient in alcoholic beverges and an octane booster in gasoline, is yielded by fermentation of glucose. The balance chemical equation is `C_(6)H_(12)O_(6)(aq.)overset("zymase")rarr 2CH_(3)CH_(2)OH(aq.)+2CO_(2)(g)` How is the rate of formation of ethanol related to the rate of consumption of glucose ?
Strategy : The rate of disappearance of a reactant X is `-Delta[X]//Delta t` while the rate of appearance of a product Y is `Delta[Y]//Delta t`. We can express the rate of the reaction in terms of either rate expressoin. To find the relative rates, look at the coefficients in the balanced chemical equation. These rate expressions can be equated , if we first divide each by the coefficient of the corresponding substance in the chemical equation.

Explanation

Rate of consumption of glucose `= - (Delta[C_(6)H_(12)O_(6)])/(Delta t)`
Rate of formation of ethanol `= (Delta[CH_(3)CH_(2)OH])/(Delta t)`
According to the balanced equation, 2 mol of ethanol are produced for each mole of glucose that reacts. Therefore, the rate of formation of ethanol is twice the rate of consumption of glucose. Divide each rate by the corresponding coefficent and then equate them : `- (Delta[C_(6)H_(12)O_(6)])/(Delta t) = (Delta[CH_(3)CH_(2)OH])/(2Delta t)`