Question

00 16t3 6. Consider the reachon Following rate nfo obtaincdl... m/S О. Ơ575 O.230 o. เ15 0-100 0.0500 Determine the rate law
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Answer #1

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

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