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For reaction `RX + OH^(-) rarr ROH + X^(-)`, rate expression is `R = 4.7 xx 10^(-5) [RX][OH^(-)] + 2.4 xx 10^(-5)[RX]`. What `%` of reactant react by `S_(N)2` mechanism when `[OH^(-)] = 0.001` molar?A. `1.9`B. `66.2`C. `95.1`D. `16.4` |
Answer» Correct Answer - a The rate is made up of two parts `S_(N^(1))` and `S_(N^(2))` `rate=4.7xx10^(-5)[RX][OH^(-)]+0.24xxunderset(S_(N^(1)))(10^(-5))[RX]` Thus `% S_(N^(2))=[S_(N^(2))/(S_(N^(2))+S_(N^(1)))]xx100` `=[(4.7xx10^(-5)[RX][OH^(-)])/(4.7xx10^(-5)[RX][OH^(-)]+0.24xx10^(-5)[RX])]xx100` `=[(4.7[OH^(-)])/(4.7[OH^(-)]+0.24)]xx100` `=(4.7xx0.001)/(4.7xx0.001+0.24)=1.9%` |
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