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(b) Cyclohexanone can be produced by the hydrogenation of phenol followed by the dissociation of the resulting cyclohexanol. \[ \begin{array}{l} 3 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}(\mathrm{~g}) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}(\mathrm{~g}) \\ \mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}(\mathrm{~g}) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \end{array} \] Under conditions which favour reaction (1), the following equilibrium partial pressures were found at 500 K . \[ \begin{array}{ll} p(\text { phenol }) & =0.625 \mathrm{~atm} \\ p(\text { cyclohexanol }) & =0.615 \mathrm{~atm} \\ p(\text { hydrogen }) & =0.200 \mathrm{~atm} \end{array} \] Under conditions which favour reaction (2), the degree of dissociation of cyclohexanol is 0.525 at 500 K and a total pressure of 2.0 atm . Calculate \( K_{p} \) for each of the reactions (1) and (2) above. An alternative synthesis proposes the replacement of the dissociation reaction (2) by reaction (3) below, between the cyclohexanol and more phenol. \[ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}(\mathrm{~g})+2 \mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}(\mathrm{~g}) \rightleftharpoons 3 \mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}(\mathrm{~g}) \] Write an expression for the equilibrium constant, \( K_{p}(3) \), for this reaction in terms of partial pressures. Express \( K_{p}(3) \) in terms of \( K_{\mathrm{p}}(1) \) and \( K_{\mathrm{p}}(2) \), and calculate \( K_{\mathrm{p}}(3) \) at 500 K .

          (b) Cyclohexanone can be produced by the hydrogenation of phenol followed by the dissociation of the resulting cyclohexanol.
\[
\begin{array}{l} 
3 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}(\mathrm{~g}) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}(\mathrm{~g}) \\
\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}(\mathrm{~g}) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})
\end{array}
\]

Under conditions which favour reaction (1), the following equilibrium partial pressures were found at 500 K .
\[
\begin{array}{ll}
p(\text { phenol }) & =0.625 \mathrm{~atm} \\
p(\text { cyclohexanol }) & =0.615 \mathrm{~atm} \\
p(\text { hydrogen }) & =0.200 \mathrm{~atm}
\end{array}
\]

Under conditions which favour reaction (2), the degree of dissociation of cyclohexanol is 0.525 at 500 K and a total pressure of 2.0 atm .
Calculate \( K_{p} \) for each of the reactions (1) and (2) above.
An alternative synthesis proposes the replacement of the dissociation reaction (2) by reaction (3) below, between the cyclohexanol and more phenol.
\[
\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}(\mathrm{~g})+2 \mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}(\mathrm{~g}) \rightleftharpoons 3 \mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}(\mathrm{~g})
\]

Write an expression for the equilibrium constant, \( K_{p}(3) \), for this reaction in terms of partial pressures. Express \( K_{p}(3) \) in terms of \( K_{\mathrm{p}}(1) \) and \( K_{\mathrm{p}}(2) \), and calculate \( K_{\mathrm{p}}(3) \) at 500 K .

(b) Cyclohexanone can be produced by the hydrogenation of phenol followed by the dissociation of the resulting cyclohexanol.

3 H2( g)+C6H5OH( g) ⇌C6H11OH( g)
C6H11OH( g) ⇌C6H10O( g)+H2( g)

Under conditions which favour reaction (1), the following equilibrium partial pressures were found at 500 K .

p( phenol ) =0.625 atm

p( cyclohexanol ) =0.615 atm

p( hydrogen ) =0.200 atm

Under conditions which favour reaction (2), the degree of dissociation of cyclohexanol is 0.525 at 500 K and a total pressure of 2.0 atm .
Calculate Kp for each of the reactions (1) and (2) above.
An alternative synthesis proposes the replacement of the dissociation reaction (2) by reaction (3) below, between the cyclohexanol and more phenol.

C6H5OH( g)+2 C6H11OH( g) ⇌ 3 C6H10O( g)

Write an expression for the equilibrium constant, Kp(3), for this reaction in terms of partial pressures. Express Kp(3) in terms of Kp(1) and Kp(2), and calculate Kp(3) at 500 K .

Added by Margaret P.

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