At $448^{\circ} \mathrm{C}$, the equilibrium constant $\left(K_C\right)$ for the reaction $\mathrm{H}_2(\mathrm{g})+\mathrm{I}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$ is $50.5 .$ If we start with $2 \times 10^{-2}$ mol of $\mathrm{HI}, 1 \times 10^{-2} \mathrm{~mol}$ of $\mathrm{H}_2$ and $3 \times 10^{-2} \mathrm{~mol} \, \mathrm{I}_2$ in a $2 \mathrm{~L}$ vessel, then at $448^{\circ} \mathrm{C}$

Reaction will proceed in the backward direction

JEE CHEMISTRY MCQ: Equilibrium

1 Medium

1 of 1

Application +4 | -1

448^{\circ} \mathrm{C}

, the equilibrium constant

\left(K_C\right)

for the reaction

\mathrm{H}_2(\mathrm{g})+\mathrm{I}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})

50.5 .

If we start with

2 \times 10^{-2}

mol of

\mathrm{HI}, 1 \times 10^{-2} \mathrm{~mol}

\mathrm{H}_2

and

3 \times 10^{-2} \mathrm{~mol} \, \mathrm{I}_2

in a

2 \mathrm{~L}

vessel, then at

448^{\circ} \mathrm{C}

View answer and explanation

Correct option: B

Reaction will proceed in the forward direction

Option B (Reaction will proceed in the forward direction) is correct because the reaction quotient (Q) is less than the equilibrium constant (K), indicating the reaction needs to shift towards products.

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.