`2NO+2H_(2)rarr N_(2)+2H_(2)O`. The experimental rate law for above reaction is , Rate `=k[NO]^(2)[H_(2)]`. When time is in minutesand the concentration is in moles `//L`, the units for `k` are

The equation and rate law for the gas phase reaction between NO and H_(2) are : 2NO(g)+2H_(2)(g)rarrN_(2)(g)+2H_(2)O(g) Rate = k[NO]^(2)[H_(2)] What are the units of k if time is in seconds and the concentration is moles per litre?

The rate law for a reaction is found to be : Rate = k [NO_2^(-) [I^(-)] [H^+]^2 How would the rate of reaction change when concentration of I^- is halved ?

For the reaction 2H_(2)(g)+2NO(g) rarr N_(2)(g)+2H_(2)O(g) the observed rate expression is, rate =k_(f)[NO]^(2)[H_(2)]. The rate experssion for the reverse reaction is :

For the reaction 2H_(2)(g)+2NO(g) rarr N_(2)(g)+2H_(2)O(g) the observed rate expression is, rate =k_(f)[NO]^(2)[H_(2)]. The rate experssion for the reverse reaction is :

For the reaction 2NO(g)+2H_(2)(g)toN_(2)(g)+2H_(2)O(g) . The rate law is rate =k[NO]^(2)[H_(2)] . What is the overall order of reaction.

For the reaction 2H_(2)+O_(2)rarr 2H_(2)O , the rate law expression is , r = k[H_(2)]^(n) . When the concentration of H_(2) is doubled, the rate of reaction found to be quadrupled. The value of n is