Degree of dissociation of an electrolyte does not depend on

  • A
    Temperature
  • B
    Concentration
  • C
    Nature of solvent
  • D
    Pressure

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For the reaction $A \rightleftharpoons \frac{1}{2} B + C$,the degree of dissociation $\alpha$ in terms of vapour density $D_t$ (theoretical) and $D_o$ (observed) is given by:

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At temperature $T$,the dissociation equilibrium of $AB_{2(g)}$ is given as: $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$. If the degree of dissociation $x$ is very small compared to $1$,then the expression for the equilibrium constant $K_p$ in terms of $x$ and total pressure $P$ is:

At $T$ $K$,consider the following gaseous reaction,which is in equilibrium: $N_2O_5 \rightleftharpoons 2NO_2 + \frac{1}{2}O_2$. What is the fraction of $N_2O_5$ decomposed at constant volume and temperature,if the initial pressure is $300 \ mm \ Hg$ and pressure at equilibrium is $480 \ mm \ Hg$? (Assume all gases as ideal)

The vapour density of $PCl_5$ is $104.16$,but when heated to $230 \, ^\circ C$,its vapour density decreases to $62$. What is the degree of dissociation of $PCl_5$ at this temperature (in $\%$) ?

Ammonia is heated in a closed vessel in the presence of a catalyst at $15 \ atm$ pressure and a temperature range of $27 \ ^\circ C$ to $347 \ ^\circ C$. Under these conditions,ammonia undergoes partial dissociation according to the reaction $2NH_3 \rightleftharpoons N_2 + 3H_2$. When the pressure is increased to $50 \ atm$ while keeping the volume of the vessel constant,calculate the percentage of $NH_3$ dissociated.

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