Question

The equilibrium constant, \( \mathrm{K}_{\mathrm{p}} \), for the following reaction is \( 1.04 \times 10^{-2} \) at \( 548 \mathrm{~K} \). \[ \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{HCl}(\mathrm{g}) \] If \( \Delta H^{\circ} \) for this reaction is \( 268 \mathrm{~kJ} \), what is the value of \( K_{p} \) at \( 448 \mathrm{~K} \) ? \[ \mathrm{K}_{\mathrm{p}}= \]

          The equilibrium constant, \( \mathrm{K}_{\mathrm{p}} \), for the following reaction is \( 1.04 \times 10^{-2} \) at \( 548 \mathrm{~K} \).
\[
\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{HCl}(\mathrm{g})
\]

If \( \Delta H^{\circ} \) for this reaction is \( 268 \mathrm{~kJ} \), what is the value of \( K_{p} \) at \( 448 \mathrm{~K} \) ?
\[
\mathrm{K}_{\mathrm{p}}=
\]

The equilibrium constant, Kp, for the following reaction is 1.04 × 10^-2 at 548 K.

NH4Cl(s) ⇌NH3( g)+HCl(g)

If Δ H^∘ for this reaction is 268 kJ, what is the value of Kp at 448 K ?

Kp=

Added by Andrew C.

Chemistry A Molecular Approach

Nivaldo J. Tro 5th Edition

Chapter 19

Instant Answer

Solved by Expert Richelle Chappell

Step 1

It is given by: \[ \ln\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H^\circ}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \] where: - \(K_1\) and \(K_2\) are the equilibrium constants at temperatures \(T_1\) and \(T_2\), respectively, - \(\Delta H^\circ\) is the Show more…

Show all steps