NEET UG CHEMISTRY MCQ: Equilibrium

1 Hard

1 of 1

Understanding +4 | -1

Consider the following reaction,

\mathrm{C}(\mathrm{s}

, diamond

)=\mathrm{C}(\mathrm{s}

, graphite

)+

heat and choose the correct option.

View answer and explanation

Correct option: B

An increase in temperature and pressure will shift the equilibrium to the left.

Option B is correct because the reaction is exothermic, and graphite is denser than diamond, so increasing temperature and pressure will shift the equilibrium to the left.

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}),

the

K_{\mathrm{P}}

value is

2.0 \mathrm{~atm}.

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

\mathrm{CaO}

and

\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

+

solvent

\leftrightharpoons

solution. For

\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