Amongst the following statements regarding adsorption, those that are valid are
(a) \(\Delta H\) becomes less negative as adsorption proceeds
(b) On a given adsorbent, ammonia is adsorbed more than nitrogen gas
(c) On adsorption, the residual force acting along the surface of the adsorbent increases
(d) With increase in temperature, the equilibrium concentration of adsorbate increases
(a) \(\Delta H\) becomes less negative as adsorption proceeds
(b) On a given adsorbent, ammonia is adsorbed more than nitrogen gas
(c) On adsorption, the residual force acting along the surface of the adsorbent increases
(d) With increase in temperature, the equilibrium concentration of adsorbate increases
Show Hint
- (c) and (d)
- (a) and (b)
- (d) and (a)
- (b) and (c)
The Correct Option is B
Solution and Explanation
(b) Easily liquefiable gases with a higher critical temperature (NH3) are adsorbed more than gases like N2 because of stronger van der Waals forces. Valid.
(c) Adsorption satisfies the unbalanced (residual) forces at the surface, so they decrease, not increase. Not valid.
(d) Adsorption is exothermic, so by Le Chatelier's principle the amount adsorbed at equilibrium decreases with rise in temperature. Not valid.
So the valid statements are (a) and (b).
Answer: (2) (a) and (b)
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Surface Chemistry Questions
- Adsorption' principle is used for which of the following purification method?
In the given TLC, the distance of spot \( A \) and \( B \) are 5 cm and 7 cm, from the bottom of the TLC plate, respectively. The \( R_f \) value of \( B \) is \( x \times 10^{-1} \) times more than \( A \). The value of \( x \) is _________ .- The adsorbent used in adsorption chromatography is/are
A. silica gel
B. alumina
C. quick lime
D. magnesia
Choose the most appropriate answer from the options given below : - The half-life of $^{65}$Zn is 245 days. After $x$ days, 75% of the original activity remained. The value of $x$ in days is __________ (Nearest integer).}
(Given: $\log 3 = 0.4771$ and $\log 2 = 0.3010$) - Match the terms in Column-I with their description in Column-II and choose the correct option:
Column I (Chemical reactions) Column II (Enzymes used) (i) Glucose → CO2 + Ethanol a Pepsin (ii) Sucrose→Glucose + Fructose b Diastase (iii) Starch →Maltose c Zymase (iv) Protein→Amino acids d Invertase
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,