228988 The equilibrium constant for the reaction, $\mathbf{a A}+\mathbf{b B} \rightleftharpoons \mathbf{c}+\mathbf{d D}$ is $\mathrm{K}$, then the equilibrium constant for the reaction, $\mathbf{n a A}+\mathbf{n b B} \rightleftharpoons \mathbf{n c C}+\mathbf{n d D}$

228989 Phosphorus pentachloride dissociates as follows in a closed reaction vessel,
$\mathbf{P C l}_{5}(\mathrm{~g}) \rightleftharpoons \mathbf{P C l}_{3}(\mathrm{~g})+\mathbf{C l}_{2}(\mathrm{~g})$
If total pressure at equilibrium of the reaction mixture is $\mathrm{p}$ and degree of dissociation of $\mathrm{PCl}_{5}$ is $\mathrm{x}$, the partial pressure of $\mathrm{PCl}_{3}$ will be

228990 For the homogeneous reaction.
$4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \rightleftharpoons 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}$
the equilibrium constant $K_{c}$ has the units

228992 For the reaction
$\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+\frac{1}{2} \mathrm{O}_{2}$
$\frac{-\mathrm{d}\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\mathrm{dt}}=\mathrm{k}_{1}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]$
$\frac{\mathrm{d}\left[\mathrm{NO}_{2}\right]}{\mathrm{dt}}=\mathrm{k}_{2}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]$
$\frac{\mathrm{d}\left[\mathrm{O}_{2}\right]}{\mathrm{dt}}=\mathrm{k}_{3}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]$

228988 The equilibrium constant for the reaction, $\mathbf{a A}+\mathbf{b B} \rightleftharpoons \mathbf{c}+\mathbf{d D}$ is $\mathrm{K}$, then the equilibrium constant for the reaction, $\mathbf{n a A}+\mathbf{n b B} \rightleftharpoons \mathbf{n c C}+\mathbf{n d D}$

228989 Phosphorus pentachloride dissociates as follows in a closed reaction vessel,
$\mathbf{P C l}_{5}(\mathrm{~g}) \rightleftharpoons \mathbf{P C l}_{3}(\mathrm{~g})+\mathbf{C l}_{2}(\mathrm{~g})$
If total pressure at equilibrium of the reaction mixture is $\mathrm{p}$ and degree of dissociation of $\mathrm{PCl}_{5}$ is $\mathrm{x}$, the partial pressure of $\mathrm{PCl}_{3}$ will be

228990 For the homogeneous reaction.
$4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \rightleftharpoons 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}$
the equilibrium constant $K_{c}$ has the units

228992 For the reaction
$\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+\frac{1}{2} \mathrm{O}_{2}$
$\frac{-\mathrm{d}\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\mathrm{dt}}=\mathrm{k}_{1}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]$
$\frac{\mathrm{d}\left[\mathrm{NO}_{2}\right]}{\mathrm{dt}}=\mathrm{k}_{2}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]$
$\frac{\mathrm{d}\left[\mathrm{O}_{2}\right]}{\mathrm{dt}}=\mathrm{k}_{3}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]$

228988 The equilibrium constant for the reaction, $\mathbf{a A}+\mathbf{b B} \rightleftharpoons \mathbf{c}+\mathbf{d D}$ is $\mathrm{K}$, then the equilibrium constant for the reaction, $\mathbf{n a A}+\mathbf{n b B} \rightleftharpoons \mathbf{n c C}+\mathbf{n d D}$

228989 Phosphorus pentachloride dissociates as follows in a closed reaction vessel,
$\mathbf{P C l}_{5}(\mathrm{~g}) \rightleftharpoons \mathbf{P C l}_{3}(\mathrm{~g})+\mathbf{C l}_{2}(\mathrm{~g})$
If total pressure at equilibrium of the reaction mixture is $\mathrm{p}$ and degree of dissociation of $\mathrm{PCl}_{5}$ is $\mathrm{x}$, the partial pressure of $\mathrm{PCl}_{3}$ will be

228990 For the homogeneous reaction.
$4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \rightleftharpoons 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}$
the equilibrium constant $K_{c}$ has the units

228992 For the reaction
$\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+\frac{1}{2} \mathrm{O}_{2}$
$\frac{-\mathrm{d}\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\mathrm{dt}}=\mathrm{k}_{1}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]$
$\frac{\mathrm{d}\left[\mathrm{NO}_{2}\right]}{\mathrm{dt}}=\mathrm{k}_{2}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]$
$\frac{\mathrm{d}\left[\mathrm{O}_{2}\right]}{\mathrm{dt}}=\mathrm{k}_{3}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]$

228988 The equilibrium constant for the reaction, $\mathbf{a A}+\mathbf{b B} \rightleftharpoons \mathbf{c}+\mathbf{d D}$ is $\mathrm{K}$, then the equilibrium constant for the reaction, $\mathbf{n a A}+\mathbf{n b B} \rightleftharpoons \mathbf{n c C}+\mathbf{n d D}$

228989 Phosphorus pentachloride dissociates as follows in a closed reaction vessel,
$\mathbf{P C l}_{5}(\mathrm{~g}) \rightleftharpoons \mathbf{P C l}_{3}(\mathrm{~g})+\mathbf{C l}_{2}(\mathrm{~g})$
If total pressure at equilibrium of the reaction mixture is $\mathrm{p}$ and degree of dissociation of $\mathrm{PCl}_{5}$ is $\mathrm{x}$, the partial pressure of $\mathrm{PCl}_{3}$ will be

228990 For the homogeneous reaction.
$4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \rightleftharpoons 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}$
the equilibrium constant $K_{c}$ has the units

228992 For the reaction
$\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+\frac{1}{2} \mathrm{O}_{2}$
$\frac{-\mathrm{d}\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\mathrm{dt}}=\mathrm{k}_{1}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]$
$\frac{\mathrm{d}\left[\mathrm{NO}_{2}\right]}{\mathrm{dt}}=\mathrm{k}_{2}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]$
$\frac{\mathrm{d}\left[\mathrm{O}_{2}\right]}{\mathrm{dt}}=\mathrm{k}_{3}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]$