Respuesta :
Answer:
The values of stoichiometric coefficients of the reaction are :
a = 1, b = 2, c = 1
Explanation:
[tex]aA+bB\rightarrow cC[/tex]
Rate of the reaction will given as:
[tex]R=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=\frac{1}{c}\frac{d[C]}{dt}[/tex]
Given :
Rate of decrease in concentration of A = [tex]-\frac{d[A]}{dt}=0.0065 mol/L s[/tex]
Rate of decrease in concentration of B = [tex]-\frac{d[B]}{dt}=0.0130 mol/L s[/tex]
[tex]-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}[/tex]
[tex]\frac{1}{a}\times 0.0065 mol/L s=\frac{1}{b}\times 0.0130 mol/L s[/tex]
[tex]a=0.5b[/tex]
b = 2a
Rate of increase in concentration of C = [tex]-\frac{d[C]}{dt}=0.0065 mol/L s[/tex]
[tex]-\frac{1}{a}\frac{d[A]}{dt}=\frac{1}{c}\frac{d[C]}{dt}[/tex]
[tex]\frac{1}{a}\times 0.0065 mol/L s=\frac{1}{c}\times 0.0065mol/L s[/tex]
a = c
[tex]aA+2aB\rightarrow aC[/tex]
[tex]A+2B\rightarrow C[/tex]
The values of stoichiometric coefficients of the reaction are :
a = 1, b = 2, c = 1
The value of coefficient in the balanced chemical equation has been, a=2, b=1, and c=1.
The general reaction for the rate has been given with the stoichiometric coefficient for the reaction.
The rate for the given reaction has been given as:
[tex]\rm Rate=\dfrac{1}{a} \dfrac{dA}{dt} \;=\;\dfrac{1}{b} \dfrac{dB}{dt} \;=\;\dfrac{1}{c} \dfrac{dC}{dt}[/tex]
Computation of the coefficient of the reaction
The value of [tex]\rm \dfrac{dA}{dt}=0.0065\;mol/L.s[/tex]
The values of [tex]\rm \dfrac{dB}{dt}=0.0130\;mol/L.s[/tex]
Substituting the values for rate:
[tex]\rm \dfrac{1}{a} \dfrac{dA}{dt}=\dfrac{1}{b} \dfrac{dB}{dt}\\\\ \dfrac{1}{a} \;\times\;0.0065= \dfrac{1}{b} \;\times\;0.0130\\\\a=0.5b\\2a=b[/tex]
Substituting b for [tex]\rm \dfrac{dC}{dt}=0.0065\;mol/L.s[/tex]
[tex]\rm- \dfrac{1}{a} \dfrac{dA}{dt}=\dfrac{1}{c} \dfrac{dC}{dt}\\\\ -\dfrac{1}{a} \;\times\;0.0065= \dfrac{1}{c} \;\times\;0.0065\\\\a=c[/tex]
From the stoichiometric:
2a = b = c
Thus, the value of coefficient in the balanced chemical equation has been, a=2, b=1, and c=1.
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