> The equilibrium constant $K_p$ at $80^\circ\text{C}$ is $1.57$ for the
> reaction:
>
> $$\ce{PH3BCl3(s) <=> PH3(g) + BCl3(g)}$$
>
> What is the minimum amount of $\ce{PH3BCl3}$ that must be placed in
> a $0.5~\text{L}$ vessel at $80^\circ\text{C}$ if equilibrium is to be attained?
Since $K_p = 1.57$, so $p_{\ce{PH3}} = p_{\ce{BCl3}} = 1.253~\text{atm}$
We also know that:
$$\frac{PV}{RT}=n$$
where $V=0.5~\text{atm}, P=1.253~\text{atm}, R=0.0821~\text{L}~\text{atm}~\text{K}^{-1}~\text{mol}^{-1}, T=353~\text{K}$.
The equilibrium constant $K_p$ at $80^{o} C$ is $1.57$ for the
reaction.
$\ce{PH_3BCl_3(s) <=> PH3(g) + BCl3(g)}$ .
What is the minimum amount of $\ce{PH_3BCl_3}$ that must be placed in
a $0.5 L$ vessel at $80^{o}C$ if equilibrium is to be attained?
Since $K_p= 1.57 \therefor...
I found two different approaches, both is giving the same answer.
Fubini:
$$
\begin{align}
\int_0^{\infty} \frac{1-e^{-ax}}{x e^x} \,dx &= \int_0^{\infty} e^{-x} \int_0^a e^{-xy} \,dy\, dx \\
&= \int_0^a \int_0^{\infty} e^{-x(1+y)}\, dx \,dy \\
&= \int_0^{a} \frac{1}{1+y}\, dy\\
&=\log (a+1) , a
