Rate law
Rate = K[ClO2]^m[OH^-]^n
0.0575 = K(0.05)^m(0.1)^n -------------------------> 1
0.230 = K(0.1)^m(0.1)^n --------------------------> 2
0.115 = K(0.1)^m(0.05)^n ----------------------> 3
equation 1 divide by equation 2
0.0575/0.23 = K(0.05)^m(0.1)^n/ K(0.1)^m(0.1)^n
0.25 = (0.05/0.1)^m
(0.5)^2 = (0.5)^m
m = 2
equation 2 divide by equation 3
0.230/0.115 = K(0.1)^m(0.1)^n/ K(0.1)^m(0.05)^n
2 = (0.1/0.05)^n
(2)^1 = (2)^n
n = 1
Rate law
Rate = K[ClO2]^2[OH^-]^1
0.0575 = K(0.05)^2(0.1)^1
K = 0.0575/(0.05)^2(0.1)^1 = 230M^-2 sec^-1
00 16t3 6. Consider the reachon Following rate nfo obtaincdl... m/S О. Ơ575 O.230 o. เ15...
Question 6 of 15 Submit A reaction has a rate law of Rate = (1.25 M-?s=')[A] [B]? What concentration of [B] would give the reaction a rate of 0.0891 if the concentration of [A] is 0.250 M?
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