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JEE CHEMISTRY MCQ: Equilibrium

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The melting points of most of the solid substances increase with an increase of pressure. However, ice melts at a temperature lower than its usual melting point when the pressure is increased. This is because:
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Correct option: A
Ice is less denser than water
Option A (Ice is less denser than water) is correct because the solid form (ice) is less dense than the liquid form (water), and increasing pressure favors the denser phase, thus lowering the melting point.

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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