Given:
Concentration of reactant used in reaction 1 = 0.130 mol/L
Concentration of reactant used in reaction 2 = 0.440 mol/L
To determine:
the rate of reaction 2 wrt to 1
Explanation:
Let the reaction be represented as:
Reactant → Product
Rate of reaction is:
[Rate] = k[reactant]
where k = rate constant
[reactant] = reactant concentration
for reactions 1 and 2 the rates are given as:
[Rate]₁ = k[reactant]₁=k[0.130] -------(1)
[Rate]₂= k[reactant]₂=k[0.440]--------(2)
2 ÷ 1
[Rate]₂/[Rate]₁ = 3.38
[Rate]₂ = 3.38[Rate]₁
Ans: Thus, reaction(2) is nearly 3 times faster than reaction(1)
Reaction 2 is 3.385 times faster compared to reaction 1
The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.
Can be formulated:
Reaction: aA ---> bB
[tex]\large{\boxed{\boxed{\bold{v~=~-\frac{\Delta A}{\Delta t}}}}[/tex]
or
[tex]\large{\boxed{\boxed{\bold{v~=~+\frac{\Delta B}{\Delta t}}}}[/tex]
A = reagent
B = product
v = reaction rate
t = reaction time
For A + B reactions ---> C + D
Reaction speed can be formulated:
[tex]\large{\boxed{\boxed{\bold{v~=~k.[A]^a[B]^b}}}[/tex]
where
v = reaction speed, M / s
k = constant, mol¹⁻⁽ᵃ⁺ᵇ⁾. L⁽ᵃ⁺ᵇ⁾⁻¹. S⁻¹
a = reaction order to A
b = reaction order to B
[A] = [B] = concentration of substances
Reaction 1 uses 0.130 mol / l of reactant, and reaction 2 uses 0.440 mol / l of reactant.
Assuming a reaction order is one then:
reaction rate 1 =
v₁ = k. [A]
v₁ = k. 0.130
reaction rate 2 =
v₂ = k. [B]
v₂ = k. 0.440, so that
[tex]\frac{v_2}{v_1}~=~k.\frac{0.440}{0.130}[/tex]
[tex]\frac{v_2}{v_1}~=~3,385[/tex]
v₂ = 3.385. v₁
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Keywords: reaction rate, reaction order, molar concentration, products, reactants
