Question

Chemical X, a powdered solid, is slowly and continuously fed for half an hour into a well-stirred vat of water. The solid quickly dissolves and hydrolyses to Y, which then slowly decomposes to Z as follows YZ,rY=kCY,k=1.5hr1\mathrm{Y} \rightarrow \mathrm{Z}, \quad-r_{\mathrm{Y}}=k C_{\mathrm{Y}}, \quad k=1.5 \mathrm{hr}^{-1} The volume of liquid in the vat stays close to 3m33 \mathrm{m}^{3} throughout this operation, and if no reaction of Y to Z occurred, the concentration of Y in the vat would be 100mol/m3100 mol/ m^{3} at the end of the half-hour addition of X. (a) What is the maximum concentration of Y in the vat and at what time is this maximum reached? (b) What is the concentration of product Z in the vat after 1 hour?

Solution

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Since Chemical XX fed to the vat of water is slowly and dissolving phenomenon is quickly its means this procedure follow zero order kinetics and in the vat, YY to ZZ is first order

Now, this series reaction is zero order followed by the first-order reaction. So here I am using results which are given in the textbook on page no. 180180

Xn1=0,k1Yn2=1k2ZrX=k1Because zero orderrY=k2CYk2=1.5hr1CX0=100molm3k1=CX0time required to pile up k1=1000.5=200molm3.hr\begin{align*} X \xrightarrow{n_{1}=0, \hskip 0.5em k_{1}}& Y \xrightarrow{n_{2}=1 \hskip 0.5em k_{2}}Z\\ -r_{X}=&k_{1} \hskip 0.5em \text{Because zero order}\\ -r_{Y}=&k_{2}C_{Y}\\ k_{2}=&1.5 \hskip 0.5em \mathrm{hr^{-1}}\\ C_{X_{0}}=&100 \hskip 0.5em \mathrm{\dfrac{mol}{m^{3}}}\\ k_{1}=&\dfrac{C_{X_{0}}}{\text{time required to pile up } }\\ k_{1}=&\dfrac{100}{0.5}=200 \hskip 0.5em \mathrm{\dfrac{mol}{m^{3}.hr}}\\ \end{align*}

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