# What is the volume after 5 hours in this first order reaction?

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If the rate constant of decomposition for #"AsH"_3(g)# is #"0.09"/"hour"# , what will be the volume of #"H"_2(g)# after #"5 hours"# , if the number of moles of #"AsH"_3(g)# is #"10 mols"# in the beginning?

If the rate constant of decomposition for

##### 1 Answer

I get about

First, write the reaction:

#2"AsH"_3(g) -> 2"As"(s) + 3"H"_2(g)#

Instead of solving for the volume of **the total volume is constant**.

Therefore, we can ignore the total volume and look at the **first-order decomposition**, we can approximate the rate law as

#r(t) = kn_("AsH"_3)# ,

where we ignore the volume of the container (even though it would normally give us a concentration if we had

We were given both the **rate constant** **initial** **s**, **initial rate**

#r(t) = ("0.09 hr"^(-1))("10 mols")#

#=# #"0.9 mols/hr"#

That means the rate of ** reaction** is

**explicitly of**

*not*

#r(t) = "0.9 mols"/"hr" = -1/2(Deltan_(AsH_3))/(Deltat)#

#= 1/2(Deltan_(As))/(Deltat) = 1/3(Deltan_(H_2))/(Deltat)#

or that

#(Deltan_(AsH_3))/(Deltat) = -"1.8 mols/hr"# .

When

#("1.8 mols AsH"_3(g))/cancel"hr" xx 5 cancel"hrs" = "9.0 mols AsH"_3(g)#

turned into products. Therefore, by stoichiometry:

#color(green)(n_("H"_2(g))) = 9.0 cancel("mols AsH"_3(g)) xx ("3 mols H"_2(g))/(2 cancel("mols AsH"_3(g)))#

#= color(green)("13.5 mols H"_2(g))#

You could also do this by directly using the rate of production of

#"0.9 mols"/"hr" = 1/3(Deltan_(H_2))/(Deltat)#

#=> (Deltan_(H_2))/(Deltat) = "2.7 mols/hr"#

#=> color(green)(n_("H"_2(g))) = ("2.7 mols H"_2(g))/cancel"hr" xx 5 cancel("hrs") = color(green)("13.5 mols H"_2(g))#

If we then assume that the reaction was occurring at

#color(blue)(V_("H"_2(g))) ~~ "22.4 L H"_2(g) xx "13.5 mols H"_2(g) = color(blue)("302.4 L")#