Question

If the rate of reaction of [0.1 M] sodium cyanide with [0.1 M] 2-bromo-2-methylpropane is 1.2 x 10-3 M/s, what would be the effect on the overall rate if the concentration of sodium cyanide is increased to [0.2 M] and the concentration of the alkyl bromide is decreased to [0.05 M]

Answer 1

Answer:

-The rate of reaction decreases by a factor of two.

Explanation:

-The initial reaction ration of Sodium Cyanide to 2-bromo-2-methylpropane is 1:1

-Increases the mole ratio of of one of the compounds by a factor of x, reduces the rate of reaction by the same factor x.

#We divide the final amount of 2-bromo-2-methylpropane  by the initial amount to obtain the increment factor,x:

$x_{NaCN}=\frac{0.2M}{0.1M}=2\\\\or\\\\x_{C_4H_9Br}=\frac{0.1m}{0.0m}=2$

Hence, the overall rate of reaction reduces by a factor of 2

Answer 2

When the concentration of sodium cyanide is increased to [0.2 M] and the concentration of the alkyl bromide is decreased to [0.05 M], the overall rate of reaction is decreased to 6 * 10^-4 M/s.

Let us recall that the rate of reaction refers to how fast or slow a reaction occurs. We should note that the  2-bromo-2-methylpropane is a tertiary alkyl halide, as such, the rate of nucleophilic substitution depends on the concentration of the alkyl halide which appears in the slow rate determining step.

Hence, when the concentration of sodium cyanide is increased to [0.2 M] and the concentration of the alkyl bromide is decreased to [0.05 M], the overall rate of reaction is decreased to 6 * 10^-4 M/s.

Learn more about alkyl halides: https://brainly.com/question/21288807