Question

At 500K in the presence of a copper surface, ethanol decomposes according to the equation

${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}\left(g\right)\mathbf{\to }{\mathbf{CH}}_{\mathbf{3}}\mathbf{CHO}\left(g\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\left(g\right)$

The presence of C₂H₅OH was measured as a function of time, and the following data were obtained

Time(s)	${\mathbf{P}}_{{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}}\left(\mathrm{torr}\right)$
0	250.
100.	237
200.	224
300.	211
400.	198
500.	185

Since the presence of a gas is directly proportional to the concentration of the gas, we can express the rate law for a gaseous reaction in terms of partial pressures. Using the preceding data, deduce the rate law, the integrated rate law, and the value of the rate constant, all in terms of process unit in atm and time in second. Predict the pressure of C₂H₅OH after 900. s from the start of the reaction.

(Hint- To determine the order of the reaction with respect to C₂H₅OH, compare now the pressure of C₂H₅OH decreases with each time listing.)

Short answer

$\mathrm{Rate}=\frac{\mathrm{dc}}{\mathrm{dt}}$

Rate constant$1.71\times {10}^{-4} \mathrm{atm}/\mathrm{s}$

The pressure of ethanol after 900s from the start of the reaction is 0.18atm.

Step-by-step solution

Finding order of reaction by using given information

• We can see in the table that for every 100second pressure is measured to decreased by 13 torr. Therefore, we can conclude that the pressure change with time is constant hence the order of reaction is zeroth order.

The rate of reaction.

$$\begin{gathered} \mathrm{Rate}=\frac{\mathrm{d}\left[\mathrm{P}\right]}{\mathrm{dt}} \\ =\frac{250-237}{100-0} \\ =0.13\mathrm{torr} \end{gathered}$$

Rate law ( for zeroth order reaction)

Rate constant=0.13 torr/s

Rate constant =$1.71\times {10}^{-4} \mathrm{atm}/\mathrm{s}$

Integrated rate law

$$\begin{gathered} \frac{dP}{dt}=K \\ {P}_t=P_{\circ }+Kt \end{gathered}$$

Pressure

$P_t=P_{\circ }+Kt$

Where P₀= Initial pressure=250 torr=$32.5\times {10}^{-2}\mathrm{atm}$

T=900s

K=${\text{1.71×10}}^{-4} \mathrm{atm}/\mathrm{s}$

$$\begin{gathered} {\mathrm{P}}_{\mathrm{t}}=1.71\times {10}^{-4} \mathrm{atm}/\mathrm{s}\times 900 \mathrm{s}+32.5\times {10}^{-2}\mathrm{atm} \\ =0.18\mathrm{atm} \end{gathered}$$

Hence pressure of ethanol after 900s from the start of the reaction is 0.18 atm.