1

JEE Advanced 2020 Paper 2 Offline

Numerical

+4

-0

An acidified solution of 0.05 M Zn

Use K

H

^{2+}is saturated with 0.1 M H_{2}S. What is the minimum molar concentration (M) of H^{+}required to prevent the precipitation of ZnS?Use K

_{sp}(ZnS) = 1.25 $$ \times $$ 10^{$$-$$22}and overall dissociation constant ofH

_{2}S, K_{net}= K_{1}K_{2}= 1 $$ \times $$ 10^{-21}.Your input ____

2

JEE Advanced 2020 Paper 1 Offline

Numerical

+4

-0

Consider the reaction,

A $$\rightleftharpoons $$ B

at 1000 K. At time t', the temperature of the system was increased to 2000 K and the system was allowed to reach equilibrium. Throughout this experiment the partial pressure of A was maintained at 1 bar. Given, below is the plot of the partial pressure of B with time. What is the ratio of the standard Gibbs energy of the reaction at 1000 K to that at 2000 K?

A $$\rightleftharpoons $$ B

at 1000 K. At time t', the temperature of the system was increased to 2000 K and the system was allowed to reach equilibrium. Throughout this experiment the partial pressure of A was maintained at 1 bar. Given, below is the plot of the partial pressure of B with time. What is the ratio of the standard Gibbs energy of the reaction at 1000 K to that at 2000 K?

Your input ____

3

JEE Advanced 2019 Paper 1 Offline

Numerical

+3

-0

For the following reaction, the equilibrium constant K

Fe

When equal volumes of

0.06 M Fe

solutions are mixed, the equilibrium concentration of Fe

_{c}at 298 K is 1.6 $$ \times $$ 10^{17}.Fe

^{2+}(aq) + S^{2-}(aq) ⇌ FeS(s)When equal volumes of

0.06 M Fe

^{2+}(aq) and 0.2 M S^{2$$ - $$}(aq)solutions are mixed, the equilibrium concentration of Fe

^{2+}(aq) is found by Y $$ \times $$ 10^{$$ - $$17}M. The value of Y is .................Your input ____

4

JEE Advanced 2018 Paper 1 Offline

Numerical

+3

-0

The solubility of a salt of weak acid $$(AB)$$ at $$pH\,$$ $$3$$ is $$Y \times {10^{ - 3}}$$ $$mol\,{L^{ - 1}}.$$ The value of $$Y$$ is ________________.

(Given that the value of solubility product of $$AB$$ $$\left( {{K_{sp}}} \right) = 2 \times {10^{ - 10}}$$ and the value of ionization constant of $$HB$$ $$\left( {{K_a}} \right) = 1 \times {10^{ - 8}}$$)

(Given that the value of solubility product of $$AB$$ $$\left( {{K_{sp}}} \right) = 2 \times {10^{ - 10}}$$ and the value of ionization constant of $$HB$$ $$\left( {{K_a}} \right) = 1 \times {10^{ - 8}}$$)

Your input ____

Questions Asked from Equilibrium (Numerical)

Number in Brackets after Paper Indicates No. of Questions

JEE Advanced 2023 Paper 1 Online (1)
JEE Advanced 2022 Paper 2 Online (1)
JEE Advanced 2022 Paper 1 Online (1)
JEE Advanced 2020 Paper 2 Offline (2)
JEE Advanced 2020 Paper 1 Offline (1)
JEE Advanced 2019 Paper 1 Offline (1)
JEE Advanced 2018 Paper 1 Offline (2)
IIT-JEE 2011 Paper 2 Offline (1)
IIT-JEE 2010 Paper 1 Offline (1)
IIT-JEE 2009 Paper 2 Offline (1)

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