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JEE CHEMISTRY MCQ: Equilibrium

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Analyzing +4 | -1
Assertion (A): At an equilibrium A(g)+2B(g)⇌C(g)\mathrm{A}(\mathrm{g})+2 \mathrm{B}(\mathrm{g}) \rightleftharpoons \mathrm{C}(\mathrm{g}) if substantial amount of water is added to the mixture and stated that only A(g)\mathrm{A}(\mathrm{g}) gets dissolved to a certain extent in water then equilibrium shifts towards forward direction.

Reason (R): On decreasing the volume of reaction mixture and keeping rest of things same the equilibrium shifts to a direction having more number of gaseous molecules.
View answer and explanation
Correct option: D
Both (A) and (R) are false
Option D is correct because both the assertion and the reason are false. Adding water and dissolving A(g) doesn't necessarily shift the equilibrium forward, and decreasing volume shifts the equilibrium towards fewer gaseous molecules, not more.

More Questions from Equilibrium

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The indicator used in titrating oxalic acid with caustic soda solution is: K1\mathrm{K}_1 and K2\mathrm{K}_2 are equilibrium constant for reactions (i) and (ii)

N2( g)+O2( g)⇌2NO(g)…\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\quad\ldots (i)

NO(g)⇌12 N2( g)+12O2( g)…\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\quad\ldots (ii)

Then, Ka1,   Ka2   and   Ka3K{a_1},\,\,\,K{a_2}\,\,\,\mathrm{and}\,\,\,K{a_3} are the respective ionisation constants for the following reactions
H2S⇋H++HS−HS−⇋H++S2−H2S⇋2H++S2−\begin{array}{l} {H_2}S\leftrightharpoons{H^ + } + H{S^ - }\\ H{S^ - }\leftrightharpoons{H^ + } + {S^{2 - }}\\ {H_2}S\leftrightharpoons2{H^ + } + {S^{2 - }} \end{array}
The correct relationship between Ka1,  Ka2 ,Ka3K{a_1},\,\,K{a_2}\,,K{a_3} is The pH\mathrm{pH} of 0.1 M−N2H40.1 \mathrm{~M}-\mathrm{N}_2 \mathrm{H}_4 solution is (For N2H4,\mathrm{N}_2 \mathrm{H}_4, Kb1=3.6×10−6,Kb2=6.4×10−12,log⁡2=0.3K_{{b}_1}=3.6 \times 10^{-6}, K_{{b}_2}=6.4 \times 10^{-12}, \log 2=0.3, log⁡3=0.48\log 3=0.48) Reaction BaO2(s)⇌BaO(s)+O2(g);ΔH=−ve.{\text{Ba}}{{\text{O}}_2}(s) \rightleftharpoons {\text{BaO}}(s) + {{\text{O}}_2}(g);\Delta H = - {\text{ve}}. In equilibrium condition pressure of O2\text{O}_2 depends on: A : At equilibrium concentration of the reactant and product does not change with time for a chemical reaction.
R: The rate of reaction is zero at equilibrium.
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