The decomposition of $\ce{AsH3}$ into solid $\ce{As}$ and $\ce{H2}$ is given in terms of pressure of the gas in a closed vessel from time to time. $$\ce{2AsH3 -> 2As + 3H2} $$ enter image description here

Show that this reaction follows first order kinetics.

My attempt :

So to prove that the reaction follows first order kinetics, I tried finding the relation between pressure in the vessel and the time taken.

I assumed the initial concentration of $\ce{AsH3}$ (which is the pressure of the whole vessel at $t=0$) as '$2a$'. Assuming the dissociation factor to be $\alpha$, at any time $t$, the composition of the vessel can be : $$\ce{AsH3} \rightarrow (2a - 2\alpha)$$ $$\ce{As} \rightarrow 2\alpha$$ $$\ce{H2} \rightarrow 3\alpha$$ (We are working with gases so the concentration of solid As can be assumed to be constant and have no effect on the pressure of the vessel)

Assuming all gases to be ideal.

So I was able to pull through till here. Assuming temperature to be constant, $a$ can found out: $$2a = {\frac{P}{RT}} \Rightarrow a = {\frac{733.22} {2 \cdot 0.081 \cdot T}} = {\frac{4465.4}{T}}$$

I am not able to move forward after this.


In a first order reaction $\ce{A \rightarrow B + C}$ with rate constant k the rate equation is $$ {\frac{d\ce{[A]}}{dt} = - k\ce{[A]} } $$ which if integrated to give the concentration vs time where the initial concentration is $\ce{[A_0]}$ produces $$\ln\left(\frac{\ce{[A]}}{\ce{[A_0]}}\right)=-kt$$ or $$ \ce{[A]= [A_0]}e^{-kt}$$

Thus a log of the ratio concentrations or pressures (or any measurement proportional to concentration) vs time should be a straight line if the reaction is first order.

In your case you are observing the product which appears as the $\ce{AsH3}$ disappears therefore $$ {\frac{d\ce{[B]}}{dt} = + k\ce{[A]} } $$

You can substitute for $\ce{[A]}$ and integrate again to get $\ce{[B]}$.

$$\ce{[B]-[B_0]=[A_0]}(1-e^{-kt})$$ However the total pressure is measured which is the sum of $\ce{2[A] + 3[B]}$ and if $\ce{[B_0] =0}$ then

$$\ce{3[B] + 2[A] =3[A_0]}(1-e^{-kt}) + \ce{2[A_0]}e^{-kt}$$ if the measured pressure is proportional to sum of concentrations with $p_0$ as initial pressure then

$$p = p_0(3-e^{-kt})$$

At $t=0$ the initial pressure is $2p_0$ which is the value given in the table. At very long times the pressure becomes constant at $3p_0$ and calculating $3p_0 - p$, rearranging and taking logs produces

$$\ln\left(\frac{3p_0 - p}{p_0}\right) = -kt $$

which when plotted is the 'normal' equation for a first order equation.

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