If we know the rate constant, can't we find the reaction rate just by looking at the coefficients?
1 Answer
For any rate law, you have:
#\mathbf(r(t) = k[A]^(m)[B]^(n)cdotcdotcdot)# for
#N# number of reactants.
Let's just say there are two for simplicity. Then...
#r(t)# is the reaction rate as a function of time, with units of#"M/s"# .#k# is the rate constant, with units of#1/("M"^(1+ m+n)cdot"s")# .#["X"]# is the concentration of reactant#"X"# .#m,n# are the orders of reactants#"A"# and#"B"# , respectively.
Even if you know
ISSUES WITH KNOWING A REACTION MECHANISM
Any given reaction can undergo multiple different mechanisms and still yield the same products.
That's how catalysts are valid choices to speed up a reaction without changing the final product(s) made.
For a general reaction
#"A" + "B" stackrel(k_("obs")" ")(>) 2"C",#
you could have a single elementary step, written as
#color(blue)("A" + "B" stackrel(k_(1)" ")(=>) 2"C").#
"
But, if we add a catalyst
#"A"# #+# #cancel("D")# #stackrel(k_(1)" ")(=>)# #cancel("E")# #+# #"C"#
#cancel("E")# #+# #"B"# #stackrel(k_(2)" ")(=>)# #"C"# #+# #cancel("D")#
#""#
#color(blue)("A")# #color(blue)(+)# #color(blue)("B")# #color(blue)(stackrel(k_("obs")" ")(>))# #color(blue)(2"C"),#
where
Notice that the mechanism is now different. However, we still get
CONCLUSIONS

In the above catalyzed reaction, the order is harder to figure out because the reaction mechanism is actually multiple steps instead of just one and you need to work with very meticulous data.

In the noncatalyzed reaction,
#k_1 = k_"obs"# , so it is easier to figure out the order based on experimental data.
That means you would need to know the details of the mechanism first before you can figure out the order.