Answer :
Answer:
Rate constant of the reaction is [tex]3.3\times 10^{-3} M^{-2} s^{-1}[/tex].
Explanation:
A + B + C → D + E
Let the balanced reaction be ;
aA + bB + cC → dD + eE
Expression of rate law of the reaction will be written as:
[tex]R=k[A]^a[B]^b[C]^c[/tex]
Rate(R) of the reaction in trail 1 ,when :
[tex][A]=0.30 M,[B]=0.30 M,[C]=0.30 M[/tex]
[tex]R=9.0\times 10^{-5} M/s[/tex]
[tex]9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c[/tex]...[1]
Rate(R) of the reaction in trail 2 ,when :
[tex][A]=0.30 M,[B]=0.30 M,[C]=0.90 M[/tex]
[tex]R=2.7\times 10^{-4} M/s[/tex]
[tex]2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c[/tex]...[2]
Rate(R) of the reaction in trail 3 ,when :
[tex][A]=0.60 M,[B]=0.30 M,[C]=0.30 M[/tex]
[tex]R=3.6\times 10^{-4} M/s[/tex]
[tex]3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c[/tex]...[3]
Rate(R) of the reaction in trail 4 ,when :
[tex][A]=0.60 M,[B]=0.60 M,[C]=0.30 M[/tex]
[tex]R=3.6\times 10^{-4} M/s[/tex]
[tex]3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c[/tex]...[4]
By [1] ÷ [2], we get value of c ;
c = 1
By [3] ÷ [4], we get value of b ;
b = 0
By [2] ÷ [3], we get value of a ;
a = 2
Rate law of reaction is :
[tex]R=k[A]^2[B]^0[C]^1[/tex]
Rate constant of the reaction = k
[tex]9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1[/tex]
[tex]k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}[/tex]
[tex]k=3.3\times 10^{-3} M^{-2} s^{-1}[/tex]