# For the reaction #2A(g) -> B(g) + C(s)#, find the rate constant at constant temperature and volume?

##
#ul("Pressure of B"" "" ""time(min)")#

#"300 Pa"" "" "" "" "10#

#"200 Pa"" "" "" "" ""completed reaction"#

##### 1 Answer

I got

**RATE LAW OF THIS KIND OF REACTION**

Recall that the **rate law** for the one-reactant, *first-order* reaction

#\mathbf(2A(g) -> B(g) + C(s))#

is written as

#\mathbf(r(t) = k[A]^m)# where:

#r(t)# is therate of reactionin#"M/min"# .#k# is therate constantin units of#1/("M"^(-1+m)cdot"min")# .#[A]# is theconcentrationof#A# in#"M"# .#m = 1# is theorderof the reactant#A# in the first-order reaction.

**RATE LAW FOR THIS PARTICULAR REACTION**

This can be rewritten as

#r(t) = kP_A,# where

#P_A# is thetotal pressurein#"Pa"# due to#A# . Remember that since#A# is the only reactant, its order is equal to the order of the reaction.

This now means the rate has units of

**THE INTEGRATED RATE LAW**

Next, we can further set the right-half of the rate law equal to the **rate of change of** **over time** (remember to include the stoichiometric coefficient as

#kP_A = -1/2(dP_A)/(dt)#

#-2kdt = 1/(P_A)dP_A#

Now, let's integrate this and see where this lands.

#-2int_(0)^(t) kdt = int_(P_(A,i))^(P_"tot") 1/(P_A)dP_A#

The **total final pressure** (upon completion) is due to only **SOLID**! Hence it does *not* contribute to the total pressure.

#stackrel("Incompletely-Derived Integrated Rate Law")overbrace(-2k(t - 0) = ln|P_"tot"| - ln|P_(A,i)|)#

The Integrated Rate Law would have been acquired by simply recognizing that

#-2kt = ln|(P_"tot")/(P_(A,i))|#

#color(green)(k = -1/(2t)ln|(P_"tot")/(P_(A,i))|)#

**DETERMINING THE TOTAL PRESSURE DUE TO A, AND THEN B**

After **10 minutes**, *partial* pressures), but after completion, *total* pressure.

Fortunately, since we were given that the reaction is at **constant temperature and volume**, that means we now can use the *mole ratio* of

From the ideal gas law, we have

#(P_(A,i)cancel(V))/(P_(B,t)cancel(V)) = (n_Acancel(RT))/(n_Bcancel(RT))#

With **reaction completion**,

We only have a *known* time for the *10-minute* trial. So, we must use the value for

**ACQUIRING THE RATE CONSTANT**

Hence, the *rate constant* at

#k = -1/(2*"10 min")ln|(P_"tot"^"10 min")/(P_(A,i))|#

#= -1/(2*"10 min")ln|"300 Pa"/"400 Pa"|#

#= (ln(4/3))/(2*"10 min")#

Assuming 3 sig figs (instead of 1) for a reasonable answer, we get:

#color(blue)(k = 1.44xx10^(-2) "min"^(-1))#

*Just goes to show you... use every piece of information from the problem if you have to. In this case, we had to.*