The rate of a chemical reaction depends on the concentrations of the reactants. let's consider the general reaction between a and b: aa+bb⇌cc+dd the dependence of the reaction rate on the concentration of each reactant is given by the equation called the rate law: rate=k[a]m[b]n where k is a proportionality constant called the rate constant. the exponent m determines the reaction order with respect to a, and n determines the reaction order with respect to b. the overall reaction order equals the sum of the exponents (m+n). for example, if m = 2 and n = 2 then rate = k[a]2 [b]2 and the overall reaction order is 2 + 2 = 4. the reaction orders are experimentally determined and usually not related to the reaction stoichiometric coefficients a and b. the reaction of ethyl bromide with sodium hydroxide, ch3ch2br(aq) + naoh(aq) ch3ch2 oh(aq) + nabr(aq) is first order in ch3ch2br and first order in naoh . if the concentration of ch3ch2br was increased by half and the concentration of naoh was quadrupled, by what factor would the reaction rate increase? express your answer numerically.

The reaction rate of ethyl bromide (CH₃CH₂Br) with sodium hydroxide (NaOH) will increase by a factor of six when the initial concentrations of ethyl bromide and sodium hydroxide are increased by half and four, respectively.  

For a reaction between A and B:

aA + bB ⇄ cC + dD

The reaction rate is proportional to the concentration of the reactants as follows:

Where:

k: is the rate constant

[A] and [B] are the concentrations of A and B, respectively

m and n are the reaction orders with respect to A and B, respectively

So, for the reaction of ethyl bromide with sodium hydroxide

CH₃CH₂Br(aq) + NaOH(aq) ⇄ CH₃CH₂OH(aq) + NaBr(aq)

we have the following rate of reaction:

Where:

i for initial

m and n are equal to 1 because the reaction is first order in CH₃CH₂Br and NaOH.

Hence, if the concentration of CH₃CH₂Br was increased by half and the concentration of NaOH was quadrupled, we have:

Then, the reaction rate is:  

Therefore, the rate will increase by a factor of six.  

You can learn more about reaction rate here:

I hope it helps you!

Ex6  planation:

CH₃CH₂Br + OH⁻ ⟶ CH₃CH₂OH + Br⁻

Let's rewrite the equation as

A + B ⟶ C + D

m = 1 and n = 1, so the rate law is

r = k[A][B]

For a first order reaction, the rate is directly proportional to the concentration.

If you increase [A] by half, you increase the rate by a factor of 1.5.

If you quadruple [B], you increase the rate by a factor of 4.

If you do both, you multiply the factors: 1.5 × 4 = 6.

The rate will increase by a factor of 6.

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Step-by-step explanation:

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