JEE CHEMISTRY MCQ: Equilibrium

1 Hard

1 of 1

Application +4 | -1

In a

10.0 \mathrm{~L}

container, an equilibrium was established between

\mathrm{SO}_3, \mathrm{SO}_2

and

\mathrm{O}_2

gases, by starting with

\mathrm{SO}_3

only. The density of equilibrium mixture was found to be

16 \mathrm{~g/litre}

at a temperature of

\frac{900}{0.0821} \mathrm{~K}.

If the degree of dissociation of

\mathrm{SO}_3

40 \%,

then the value of

K_{{P}}

(in atm) for the reaction

2 \mathrm{SO}_3(\mathrm{g})\rightleftharpoons 2 \mathrm{SO}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g})

View answer and explanation

Correct option: A

16 \mathrm{~atm}

The correct answer is 16 atm. The equilibrium constant

K_P

is calculated using partial pressures derived from the degree of dissociation and total pressure.

More Questions from Equilibrium

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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