The reaction, AB => A(g) + B(g), has a rate law, rate =k[AB] 2 and k = 0.015 L/m

Question

The reaction, AB => A(g) + B(g), has a rate law, rate =k[AB]2 and k = 0.015 L/mol.sec. What will be theconcentration of [AB] after 61 seconds from an initialconcentration of 0.408 mol/L. (Use scientific notation using theE-format. e.g., 0.0111 is 1.11E-2 Please help! The reaction, AB => A(g) + B(g), has a rate law, rate =k[AB]2 and k = 0.015 L/mol.sec. What will be theconcentration of [AB] after 61 seconds from an initialconcentration of 0.408 mol/L. (Use scientific notation using theE-format. e.g., 0.0111 is 1.11E-2 Please help!

Explanation / Answer

Since rate = k[AB]2 So it is a second order reactionso the rate constant isdefined as K = x / ( ta(a-x)) Where a = initial concentration = 0.408 mol / L          ( a-x ) =concentration left after time t = ? x = a - ( a-x) = 0.408 - ( a-x)     t = time taken = 61 s K = rate constant = 0.015 L / mol sec Plug the values we get K = x / ( ta(a-x))                           0.015 = 0.408 - ( a-x) / [ 61 * 0.408 * ( a-x) ]                          ==> (a-x ) = 0.297 mol / L                                             =2.97E-1 So the required concentration is 2.97E-1

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