Embibe Experts Solutions for Chapter: Equilibrium, Exercise 1: Exercise 1

In a $0.25$ litre tube, dissociation of $4$ moles of $\mathrm{NO}$ takes place. If its degree of dissociation is $10\%$, the value of ${\mathrm{K}}_{\mathrm{p}}$ for reaction $\begin{array}{cccc}2\mathrm{NO} \left(\mathrm{g}\right)& \rightleftharpoons & {\mathrm{N}}_{2 }\left(\mathrm{g}\right)+& {\mathrm{O}}_2\end{array}\left(\mathrm{g}\right)$ is:

$4$ mole of $\mathrm{A}$ are mixed with $4$ mole of $\mathrm{B}$, when $2$ mole of $\mathrm{C}$ are formed at equilibrium, according to the reaction. $\mathrm{A}+\mathrm{B}\rightleftharpoons \mathrm{C}+\mathrm{D}$. The equilibrium constant is:

A mixture of $0.3$ mole of ${\mathrm{H}}_2$ and $0.3$ mole of ${\mathrm{I}}_2$ is allowed to react in a $10$ litre evacuated flask at $500°\mathrm{C}$. The reaction is ${\mathrm{H}}_2+{\mathrm{I}}_2\rightleftharpoons 2\mathrm{HI}$, the $\mathrm{K}$ is found to be $64$. The amount of unreacted ${\mathrm{I}}_2$ at equilibrium is 

$\mathrm{pH}$ of human blood is $7.4$. Then ${\mathrm{H}}^{+}$ concentration will be

The ionic product of water at $25°\mathrm{C}$ is ${10}^{-14}$. The ionic product at $90°\mathrm{C}$ will be

A buffer solution with $\mathrm{pH} 9$ is to be prepared by mixing ${\mathrm{NH}}_4\mathrm{Cl}$ and ${\mathrm{NH}}_4\mathrm{OH}$. Calculate the number of moles of ${\mathrm{NH}}_4\mathrm{Cl}$ that should be added to one litre of $1.0 \mathrm{M} {\mathrm{NH}}_4\mathrm{OH}. \left[{\mathrm{K}}_{\mathrm{b}}=1.8\times {10}^{-5}\right]$

Which is the correct representation of the solubility product constant of ${\mathrm{Ag}}_2{\mathrm{CrO}}_4$

The effect of temperature on the equilibrium constant is expressed as, $\left({\mathrm{T}}_2>{\mathrm{T}}_1\right) \log \left(\frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}\right)=\frac{-\mathrm{\Delta H}}{2.303\mathrm{R}}\left[\frac{1}{ {\mathrm{T}}_2}-\frac{1}{ {\mathrm{T}}_1}\right]$. For endothermic reactions, false statement is: