Question

The equilibrium constant, \( K_{p} \), for the following reaction is \( \mathbf{0 . 6 3 6} \) at \( 600 \mathrm{~K} \). \[ \mathrm{COCl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}_{(\mathrm{g})}+\mathrm{Cl}_{2}(\mathrm{~g}) \] If \( \Delta H^{\circ} \) for this reaction is \( 108 \mathrm{~kJ} \), what is the value of \( \mathrm{K}_{\mathrm{p}} \) at \( 693 \mathrm{~K} \) ? \[ K_{p}= \]

          The equilibrium constant, \( K_{p} \), for the following reaction is \( \mathbf{0 . 6 3 6} \) at \( 600 \mathrm{~K} \).
\[
\mathrm{COCl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}_{(\mathrm{g})}+\mathrm{Cl}_{2}(\mathrm{~g})
\]

If \( \Delta H^{\circ} \) for this reaction is \( 108 \mathrm{~kJ} \), what is the value of \( \mathrm{K}_{\mathrm{p}} \) at \( 693 \mathrm{~K} \) ?
\[
K_{p}=
\]

The equilibrium constant, Kp, for the following reaction is 0 . 6 3 6 at 600 K.

COCl2( g) ⇌CO(g)+Cl2( g)

If Δ H^∘ for this reaction is 108 kJ, what is the value of Kp at 693 K ?

Kp=

Added by Andrew C.

General Chemistry

Darrell Ebbing, Steven D. Gammon 11th Edition

Chapter 14

Instant Answer

Solved by Expert Ishu Khandelwal

Step 1

The equation is given by: \[ \ln\left(\frac{K_{p2}}{K_{p1}}\right) = \frac{-\Delta H^{\circ}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \] where - \( K_{p1} \) is the equilibrium constant at the initial temperature \( T_1 \), - \( K_{p2} \) is the equilibrium Show more…

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