CH4(g) + O2(g) ?? CO2(g) + H2O(g) The rate equality for the unbalanced reaction

Question

CH4(g) + O2(g) ?? CO2(g) + H2O(g)
The rate equality for the unbalanced reaction Ithought to be: Rate = [CH4+]m/t=[O2-]n/t with the rate law: Rate =-K[CH4+]m[O2-]n Calculate the averate rate if initially the reactantconcentraton is 2.0M and after 10.0 seconds the reactantconcentration is 0.50M I'm confused because I have 10.0 seconds but starting with 0?for initially? I thought possibly: (0.50m-2.0m)/(10s-0s) = -0.15 m/s is thiscorrect? CH4(g) + O2(g) ?? CO2(g) + H2O(g)
The rate equality for the unbalanced reaction Ithought to be: Rate = [CH4+]m/t=[O2-]n/t with the rate law: Rate =-K[CH4+]m[O2-]n Calculate the averate rate if initially the reactantconcentraton is 2.0M and after 10.0 seconds the reactantconcentration is 0.50M I'm confused because I have 10.0 seconds but starting with 0?for initially? I thought possibly: (0.50m-2.0m)/(10s-0s) = -0.15 m/s is thiscorrect?

Explanation / Answer

         CH4(g) + O2(g)?? CO2(g) +H2O(g)                 Averagerate of the reaction = Change in concentration / change intime                                                       = (0.5 M - 2.0M) / (10s - 0s)                                                       = -0.15 M /s         The negativesign indicates the reactant concentration decrease with time.                rate= - dX /dt = 0.15 M/s

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