\(2NOCl(g) ⇌ 2NO(g) + Cl_2(g)\)
\(t = 0\) \(2\) \(-\) \(-\)
\(t = t_{eq}\) \((2-0.4=1.6)\) \(0.4\) \(0.2\)
Equilibrium constant,
\(k_c = \frac {(0.2)\times(0.4)^2}{(1.6)^2}\)
\(k_c = \frac {0.2}{16}\)
\(k_c = \frac 18 \times10^{-1}\)
\(k_c = 0.125 \times 10^{-1}\)
\(k_c = 125 \times 10^{-4}\)
Let given that the equilibrium constant at \(30°C\) is \(x×10^{–4}.\)
Then, \(x = 125\)
So, the answer is \(125\).
A force of \( F = 0.5 \) N is applied on the lower block as shown in the figure. The work done by the lower block on the upper block for a displacement of 3 m of the upper block with respect to the ground is (Take, \( g = 10 \) m/s\( ^2 \)):
A molecule with the formula $ \text{A} \text{X}_2 \text{Y}_2 $ has all it's elements from p-block. Element A is rarest, monotomic, non-radioactive from its group and has the lowest ionization energy value among X and Y. Elements X and Y have first and second highest electronegativity values respectively among all the known elements. The shape of the molecule is:
A transition metal (M) among Mn, Cr, Co, and Fe has the highest standard electrode potential $ M^{n}/M^{n+1} $. It forms a metal complex of the type $[M \text{CN}]^{n+}$. The number of electrons present in the $ e $-orbital of the complex is ... ...
Consider the following electrochemical cell at standard condition. $$ \text{Au(s) | QH}_2\text{ | QH}_X(0.01 M) \, \text{| Ag(1M) | Ag(s) } \, E_{\text{cell}} = +0.4V $$ The couple QH/Q represents quinhydrone electrode, the half cell reaction is given below: $$ \text{QH}_2 \rightarrow \text{Q} + 2e^- + 2H^+ \, E^\circ_{\text{QH}/\text{Q}} = +0.7V $$
0.1 mol of the following given antiviral compound (P) will weigh .........x $ 10^{-1} $ g.
Consider the following equilibrium, $$ \text{CO(g)} + \text{H}_2\text{(g)} \rightleftharpoons \text{CH}_3\text{OH(g)} $$ 0.1 mol of CO along with a catalyst is present in a 2 dm$^3$ flask maintained at 500 K. Hydrogen is introduced into the flask until the pressure is 5 bar and 0.04 mol of CH$_3$OH is formed. The $ K_p $ is ...... x $ 10^7 $ (nearest integer).
Given: $ R = 0.08 \, \text{dm}^3 \, \text{bar} \, \text{K}^{-1} \, \text{mol}^{-1} $
Assume only methanol is formed as the product and the system follows ideal gas behavior.
An equilibrium represents a state in a process when the observable properties such as color, temperature, pressure, concentration etc do not show any change.
The word equilibrium means ‘balance’ which indicates that a chemical reaction represents a balance between the reactants and products taking part in the reaction. The equilibrium state is also noticed in certain physical processes such as the melting point of ice at 0℃, both ice and water are present at equilibrium.
In the case of physical processes such as the melting of solid, dissolution of salt in water etc., the equilibrium is called physical equilibrium while the equilibrium associated with chemical reaction is known as chemical equilibrium.
The chemical equilibrium in a reversible reaction is the state at which both forward and backward reactions occur at the same speed.
The stage of the reversible reaction at which the concentration of the reactants and products do not change with time is called the equilibrium state.
Read More: Calculating Equilibrium Concentration
There are two types of chemical equilibrium:
In this type, the reactants and the products of chemical equilibrium are all in the same phase. Homogenous equilibrium can be further divided into two types: Reactions in which the number of molecules of the products is equal to the number of molecules of the reactants. For example,
Reactions in which the number of molecules of the products is not equal to the total number of reactant molecules. For example,
In this type, the reactants and the products of chemical equilibrium are present in different phases. A few examples of heterogeneous equilibrium are listed below.
Thus, the different types of chemical equilibrium are based on the phase of the reactants and products.
Check Out: Equilibrium Important Questions