For the reaction, #"2XO + O"_2 → "2XO"_2#, data obtained from measurement of the initial rate of reaction at varying concentrations are?

Question

#ulbb("Experiment"color(white)(m)["XO"]color(white)(m) ["O"_2]color(white)(m)"Rate")#
#color(white)(mmm)1color(white)(mmmml) 0.010color(white)(m) 0.010color(white)(mll) 2.5#
#color(white)(mmm)2color(white)(mmmml) 0.010color(white)(m) 0.020color(white)(mll) 5.0#
#color(white)(mmm)3color(white)(mmmml) 0.030color(white)(m) 0.020color(white)(ml) 45.0#

What is the rate law for this reaction?

#"rate" = k["XO"]^2/["O"_2]^2#

#"rate" = k["XO"]["O"_2]#

#"rate" = k["XO"]["O"_2]^2#

#"rate" = k["XO"]^2 ["O"_2]^2#

#"rate" = k["XO"]^2 ["O"_2]#

Naomi Aronson · Accepted Answer

The rate law is 5. #r = k["XO"]^2["O"_2]#.

The rate law is #r = k"[XO]"^m["O"_2]"^n# where #rcolor(white)(mll) =# the reaction rate #k color(white)(mll)=# the rate constant #m, n =# integers whose value we must determine Step 1. The order with respect to #"[XO]"# We must find two experiments in which only #"[XO]"# changes but #["O"_2]# stays constant. Consider experiments 2 and 3. #r_3/r_2 = (color(red)(cancel(color(black)(k)))0.030 ^mcolor(red)(cancel(color(black)(0.020^n))))/(color(red)(cancel(color(black)(k)))0.010 ^mcolor(red)(cancel(color(black)(0.020^n)))) = 45.0/5.0 # #3.0^m = 9.0# Thus, tripling the concentration multiplies the rate by nine. The reaction is second order in #["XO"]#. If you use logarithms, you get #color(blue)(mlog3.0 = log9.0)# #color(blue)(m = log9.0/log3.0 = 2.00 ≈ 2)# The reaction is second order in #"[XO]"#. Step 2. The order with respect to #["O"_2]# We must find two experiments in which only #["O"_2]# changes but #"[XO]"# stays constant. Consider experiments 1 and 2. #r_2/r_1 = (color(red)(cancel(color(black)(k)))color(red)(cancel(color(black)(0.010 ^m)))0.020^n)/(color(red)(cancel(color(black)(k)))color(red)(cancel(color(black)(0.010 ^m)))0.010^n) = 5.0 /2.5# #2.0^n = 2.0# Doubling the concentration doubles the rate, The reaction is first order in #["O"_2]# If you use logarithms, you get #color(blue)(mlog2.0 = log2.0)# #color(blue)(m = log2.0/log2.0 = 1.00 ≈ 1)# The reaction is first order in #["O"_2]#. The rate law is #r = k["XO"]^2["O"_2]#.

Serenity Allen · Answer

undefined To provide a comprehensive answer, I would need specific data on the initial rates of reaction at varying concentrations for the given reaction "2XO + O2 → 2XO2". This typically involves experimental data obtained through measurements of the rate of reaction under different initial concentrations of the reactants.

Without the specific data, it's not possible to give a detailed response. However, typically, the initial rate of reaction data would include measurements of the initial concentrations of the reactants (e.g., [XO], [O2]), the corresponding initial rate of reaction, and potentially the units in which these quantities are measured. This data is then used to determine the rate equation and rate constant for the reaction.