In a $5.76 \mathrm{~L}$ vessel, 0.5 moles of $\mathrm{H}_2$ gas and 0.5 moles of $\mathrm{I}_2$ vapours are allowed to react to form $\mathrm{HI}(\mathrm{g})$ at $447^{\circ} \mathrm{C},$ then the total pressure of gases at equilibrium would be $(R=0.08 \mathrm{~L}-\mathrm{atm} / \mathrm{K}-\mathrm{mol})$

JEE CHEMISTRY MCQ: Equilibrium

1 Medium

1 of 1

Application +4 | -1

In a

5.76 \mathrm{~L}

vessel, 0.5 moles of

\mathrm{H}_2

gas and 0.5 moles of

\mathrm{I}_2

vapours are allowed to react to form

\mathrm{HI}(\mathrm{g})

447^{\circ} \mathrm{C},

then the total pressure of gases at equilibrium would be

(R=0.08 \mathrm{~L}-\mathrm{atm} / \mathrm{K}-\mathrm{mol})

View answer and explanation

Correct option: B

10 \mathrm{~atm}

Option B (10 atm) is correct because the total pressure at equilibrium is calculated using the ideal gas law, considering that the number of moles remains constant due to

\Delta n_g = 0

More Questions from Equilibrium

Practice All Chapter MCQs →

The indicator used in titrating oxalic acid with caustic soda solution is:

\mathrm{K}_1

and

\mathrm{K}_2

are equilibrium constant for reactions (i) and (ii)

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

(i)

\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,

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

are the respective ionisation constants for the following reactions

\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

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

is The

\mathrm{pH}

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

solution is (For

\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

) Reaction

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

In equilibrium condition pressure of

\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.