Question

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.

Answer 1

Answer:

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.

Answer 2

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:

$rate = k[A]^{m}[B]^{n}$  

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:

$rate_{i} = k[CH_{3}CH_{2}Br]_{i}[NaOH]_{i}$

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:

$[CH_{3}CH_{2}Br]_{f} = 0.5[CH_{3}CH_{2}Br]_{i} + [CH_{3}CH_{2}Br]_{i} = 1.5[CH_{3}CH_{2}Br]_{i}$

$[NaOH]_{f} = 4[NaOH]_{i}$

Then, the reaction rate is:  

$rate_{f} = k[CH_{3}CH_{2}Br]_{f}[NaOH]_{f} = (1.5[CH_{3}CH_{2}Br]_{i})(4[NaOH]_{i}) = 6rate_{i}$

             

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

     

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