19,000+ solved questions for JEE Advanced, JEE Mains, NEET & IChO — with answers and expert explanations.
N2O4 2NO2, Kc = 4. This reversible reaction is studied graphically as shown in the given figure. Select the correct statement out of I, II and III. I : Reaction quotient has maximum value at point A II : Reaction proc…
I I I This reaction is set-up in aqueous medium. We start with 1 mol of I2 and 0.5 mol of I– in 1L flask. After equilibrium is reached, excess of AgNO3 gave 0.25 mol of yellow ppt. Equilibrium constant is : [Given…
When A2 and B2 are allowed to react, the equilibrium constant of the reaction at 27ºC is found (Kc = 4). A2(g) + B2(g) 2AB(g) What will be the equilibrium concentration of AB ?
Consider the reaction NaBr(aq) + H2SO4(aq) NaHSO4(aq) + HBr(aq). The equilibrium constant is 8.3 × 10 –2. If the equilibrium concentrations of NaBr, H2SO4 and NaHSO4 are 6.0 M, 9.0 M and 3.0 M respectively, what is th…
For the all gas reaction at 1000 K : 2HI H2 + I2, K = 0.0344. If you were to begin with 8.0 M of HI in a fixed volume container, what would be the equilibrium concentration of H2 ?
In reversible reaction k k A B , the initial concentration of A and B are a and b in moles per litre and the equilibrium concentration are (a – x) and (b + x) respectively ; express x in terms of k1, k2, a and b. (a…
If for the heterogeneous equilibrium CaCO3(s) CaO(s) + CO2 (g) ; K = 1 at 1 atm, the temperature is given by : (a) Sº T Hº (b) Hº T Sº (c) Gº T R (d) Gº T Hº
In each of the following, total pressure set-up at equilibrium is assumed to be equal and is one atm with equilibrium constants KP given : I :CaCO3 (s) CaO (s) + CO2 (g), K1 II : NH4HS (s) NH3 (g) + H2S (g), K2 III :…
Variation of log K with 1 T is shown by the following graph. Hº for the reaction will be :
In a 1.0 L aqueous solution when the reaction (aq) (s) 2 Ag Cu (aq) (s) Cu 2 Ag reaches equilibrium, [Cu2+] = x M and [Ag+] = y M. If the volume of solution is doubled by adding water, then at equilibrium : (a) x …
Assertion : For a reaction at equilibrium, the free energy for the reaction is minimum. Reason : The free energy for both reactants and products decreases and become equal.
Assertion : If some PCl5(g) containing labelled phosphorus P31 is added to a system with following equilibrium, after sometime the system was found to contain radioactive PCl3. PCl (g) PCl (g) Cl (g) Reason : Chemic…
Assertion : For a reversible reaction, Gibbs’ free energy is zero at equilibrium. Reason : As reaction approaches equilibrium, Gibbs’ free energy decreases.
Assertion : If the equation for a reaction is reversed, the equilibrium constant is inverted and if the equation is multiplied by 2, the equilibrium constant is squared. Reason : The numerical value of equilibrium con…
Assertion : Kp is equal to Kc for the following reaction: N2(g) + O2(g) 2NO(g) Reason : Kp and Kc are interrelated by the following equation : g n p c K K (RT) ng = Number of gaseous moles of products – Number of g…
Assertion : For the reaction g g g g A B C D at the given temperature, there will be no effect by addition of inert gas either at constant pressure or at constant volume. Reason : For the reaction where n = 0,…
Assertion : Addition of an inert gas at constant pressure equilibrium will support the dissociation of PCl5 at the same temperature. Reason : The addition of an inert gas will tend to increase the total pressure.
Assertion : The addition of an inert gas under constant volume condition cause the total pressure of the system to increase. This would thus lower the degree of dissociation of PCl5. Reason : Addition of inert gas at …
Assertion :A catalyst does not influence the value of equilibrium constant. Reason : Catalyst influence the rate of both forward and backward reactions equally.
Assertion : For PCl5(g) PCl3(g) + Cl2(g). If more Cl2 is added, the equilibrium will shift in backward direction hence equilibrium constant will decrease. Reason : Addition of inert gas to the equilibrium mixture at…
If standard heat of dissociation of PCl5 is 230 cal, then slope of the graph of log k vs 1 T is
For exothermic reaction is Sº < 0, then the sketch of log K vs 1/T may be (a) (b) (c) (d)
If for a particular reversible reaction : Kc = 57 at 355ºC and Kc = 69 at 450º, then
When a gas “X” is dissolved in water, the heat is evolved. Then solubility of ‘X’ will increase at
Au(s) Au(l) Above equilibrium is favoured at