Question #45ea5

Question

2051 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-12T18:09:13

1 hour

If the reaction is zero order we can write:

#sf("Rate"=k[NH_3]^0=k)#

#sf(k=1.50xx10^(-6)color(white)(x)"M/s")#

This means that #sf(1.50xx10^(-6))# moles per litre is being lost every second.

The total loss in concentration #sf(=(6.40-1.00)10^(-3)=5.40xx10^(-3)" ""M")#

#:.# Time taken = #sf((5.40xx10^-3)/(1.50xx10^-6)# #sf((M)/("M/s")=3.6xx10^3" "s)#

#sf(=60color(white)(x)"min")#

#sf(=1color(white)(x)"hr")#