Class 11Chemistry6-1.Equilibrium (Chemical Equilibrium)Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)
MediumFor the reaction $CH_4(g) + 2O_2(g) \rightleftharpoons CO_2(g) + 2H_2O(l)$,the enthalpy change is $\Delta H_r = -170.8 \, kJ \, mol^{-1}$. Which of the following statements is $NOT$ true?
- AAt equilibrium,the concentrations of $CO_2(g)$ and $H_2O(l)$ are not equal.
- BThe equilibrium constant for the reaction is given by $K_p = \frac{P_{CO_2}}{P_{CH_4} \cdot P_{O_2}^2}$.
- CAdding $CH_4(g)$ or $O_2(g)$ at equilibrium shifts the reaction to the right.
- DThe reaction is exothermic.
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Similar Questions
$2$ moles of $N_2$ are mixed with $6$ moles of $H_2$ in a closed vessel of $1 \ L$ capacity. If $50\%$ of $N_2$ is converted into $NH_3$ at equilibrium,find the value of $K_c$ for the reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
Medium
View SolutionIn a reaction,$A + B \rightleftharpoons C + D$,$40 \%$ of $B$ has reacted at equilibrium,when $1 \ mol$ of $A$ was heated with $1 \ mol$ of $B$ in a $10 \ L$ closed vessel. The value of $K_C$ is
Which of the following statements is/are wrong?
$(a)$ At equilibrium,concentrations of reactants and products become constant because the reaction stops.
$(b)$ Addition of catalyst speeds up the forward reaction more than the backward reaction.
$(c)$ Equilibrium constant of an exothermic reaction decreases with increase of temperature.
$(d)$ $K_p$ is always greater than $K_c$.
$(a)$ At equilibrium,concentrations of reactants and products become constant because the reaction stops.
$(b)$ Addition of catalyst speeds up the forward reaction more than the backward reaction.
$(c)$ Equilibrium constant of an exothermic reaction decreases with increase of temperature.
$(d)$ $K_p$ is always greater than $K_c$.
Medium
View Solution$56 \, g$ of nitrogen and $8 \, g$ of hydrogen gas are heated in a closed vessel. At equilibrium,$34 \, g$ of ammonia are present. The equilibrium number of moles of nitrogen,hydrogen,and ammonia are respectively:
Medium
View SolutionAt $67\,^{\circ}C$ and $1\ bar$ pressure,dinitrogen tetraoxide is $50\%$ dissociated into nitrogen dioxide. $\Delta G^{\circ}$ for the process $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ is $(R = \frac{25}{3} \ J \ K^{-1} \ mol^{-1}, \ln 2 = 0.7, \ln 3 = 1.1)$.
Medium
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