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NEET UG CHEMISTRY MCQ: Equilibrium

1 Easy

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Application +4 | -1

Densities of diamond and graphite are

3.5

and

2.3\;\mathrm{g}/\mathrm{mL}

respectively. Increase of pressure on the equilibrium

{\mathrm{C}_{\mathrm{diamond}}}\leftrightharpoons{\mathrm{C}_{\mathrm{graphite}}}

View answer and explanation

Correct option: C

Have no effect

The correct answer is C (Have no effect) because pressure favors the phase with smaller volume. Diamond has higher density hence smaller volume, so increasing pressure will favor diamond formation, but the question asks about the equilibrium, so the change in pressure will shift the equilibrium but not affect it.

More Questions from Equilibrium

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Given below are two statements: One is labelled as Assertion A and the other labelled as Reason R.
Assertion A: During the boiling of water having temporary hardness,

Mg{(HC{O_3})_2}

is converted to

MgC{O_3}

.
Reason R The solubility product of

Mg{(OH)_2}

is greater than that of

MgC{O_3}

.
In the light of the above statements, choose the most appropriate answer from the options given below At a certain temperature,

K_{\mathrm{P}}

for the dissociation of solid

\mathrm{CaCO}_3

4.5 \times 10^{-2}

atm and for the reaction

\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}),

K_{\mathrm{P}}

2.0 \mathrm{~atm}.

The pressure of carbon monoxide at this temperature, when solid carbon

\mathrm{CaO}

\mathrm{CaCO}_3

are mixed together and allowed to attain equilibrium is Phenolphthalein shows … in acid medium. At the state of dynamic equilibrium, for solute

+

\leftrightharpoons

\mathrm{PC}{\mathrm{l}_5}\leftrightharpoons\mathrm{PC}{\mathrm{l}_3} + \mathrm{C}{\mathrm{l}_2}

, initial concentration of each reactant and product is 1 M. If

{K_{\mathrm{eq}}} = 0.41

then Select the

\mathrm{p}{K_a}

value of the strongest acid from the following