The following reactions are known to occur in the body, ${\rm{C}}{{\rm{O}}_2} + {{\rm{H}}_2}{\rm{O}}\leftrightharpoons{{\rm{H}}_2}{\rm{C}}{{\rm{O}}_3}\leftrightharpoons{{\rm{H}}^ + } + {\rm{HCO}}_3^ - $ If ${\rm{C}}{{\rm{O}}_2}$ escapes from the system, then:

${{\rm{H}}_2}{\rm{C}}{{\rm{O}}_3}$ concentration will be unaltered

The forward reaction will be promoted

JEE CHEMISTRY MCQ: Equilibrium

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The following reactions are known to occur in the body,

\mathrm{C}{\mathrm{O}_2} + {\mathrm{H}_2}\mathrm{O}\leftrightharpoons{\mathrm{H}_2}\mathrm{C}{\mathrm{O}_3}\leftrightharpoons{\mathrm{H}^ + } + \mathrm{HCO}_3^ -

\mathrm{C}{\mathrm{O}_2}

escapes from the system, then:

View answer and explanation

Correct option: B

Hydrogen ion concentration will diminish

Option B (Hydrogen ion concentration will diminish) is correct because if

CO_2

escapes, the equilibrium shifts left, reducing

H_2CO_3

and therefore

H^+

concentration.

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\mathrm{K}_1

\mathrm{K}_2

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

\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\quad\ldots

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

\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}

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

\mathrm{pH}

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

\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

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

\text{O}_2