JEE CHEMISTRY MCQ: Equilibrium

1 Hard

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Application +4 | -1

Consider the following equilibrium at

25^{\circ} \mathrm{C}

\mathrm{CuSO}_4.5 \mathrm{H}_2 \mathrm{O}{(\mathrm{s})} \rightleftharpoons \mathrm{CuSO}_4.3 \mathrm{H}_2 \mathrm{O}{(\mathrm{s})}+2 \mathrm{H}_2 \mathrm{O}{(\mathrm{g})}

If this reaction takes place at two cities

\mathrm{A}

and

\mathrm{B}

having relative humidity

60 \%

and

80 \%

respectively. The vapour pressure of water at

25^{\circ} \mathrm{C}

20 \mathrm{~mm} \mathrm{~Hg}

and

\mathrm{K}_{\mathrm{P}}

for the above reaction is

4 \times 10^{-2}(\mathrm{atm})^2

25^{\circ} \mathrm{C}

then

\mathrm{CuSO}_4.5 \mathrm{H}_2 \mathrm{O}

acts as efflorescent in which city?

View answer and explanation

Correct option: A

At A only

Option A (At A only) is correct because CuSO4.5H2O acts as efflorescent when the partial pressure of water vapor in the atmosphere is less than the equilibrium vapor pressure, which is the case in city A.

More Questions from Equilibrium

Practice All Chapter MCQs →

The indicator used in titrating oxalic acid with caustic soda solution is:

\mathrm{K}_1

and

\mathrm{K}_2

are equilibrium constant for reactions (i) and (ii)

\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\quad\ldots

(i)

\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,

K{a_1},\,\,\,K{a_2}\,\,\,\mathrm{and}\,\,\,K{a_3}

are the respective ionisation constants for the following reactions

\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

K{a_1},\,\,K{a_2}\,,K{a_3}

is The

\mathrm{pH}

0.1 \mathrm{~M}-\mathrm{N}_2 \mathrm{H}_4

solution is (For

\mathrm{N}_2 \mathrm{H}_4,

K_{{b}_1}=3.6 \times 10^{-6}, K_{{b}_2}=6.4 \times 10^{-12}, \log 2=0.3

\log 3=0.48

) Reaction

{\text{Ba}}{{\text{O}}_2}(s) \rightleftharpoons {\text{BaO}}(s) + {{\text{O}}_2}(g);\Delta H = - {\text{ve}}.

In equilibrium condition pressure of

\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.