Class 11Chemistry6-1.Equilibrium (Chemical Equilibrium)Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)
MediumFor the reaction,$2SO_{2(g)} + O_{2(g)} \to 2SO_{3(g)}$
$\Delta H = -57.2 \ kJ \ mol^{-1}$ and $K_C = 1.7 \times 10^{16}$
Which of the following statement is $INCORRECT$?
$\Delta H = -57.2 \ kJ \ mol^{-1}$ and $K_C = 1.7 \times 10^{16}$
Which of the following statement is $INCORRECT$?
- AThe equilibrium constant is large suggestive of reaction going to completion and so no catalyst is required.
- BThe equilibrium will shift in forward direction as the pressure increases.
- CThe equilibrium constant decreases as the temperature increases.
- DThe addition of inert gas at constant volume will not affect the equilibrium constant.
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Similar Questions
In reaction $A + 2B \rightleftharpoons 2C + D$,the initial concentration of $B$ was $1.5$ times that of $[A]$,but at equilibrium,the concentrations of $A$ and $B$ became equal. The equilibrium constant for the reaction is:
DifficultJEE MAIN 2019
View Solution$CH_3COCH_{3(g)} \rightleftharpoons C_2H_{6(g)} + CO_{(g)}$. The initial pressure of $CH_3COCH_3$ is $100 \ mm$. When equilibrium is set up,the mole fraction of $CO_{(g)}$ is $\frac{1}{4}$. Hence,the partial pressure of $CO$ is:
Medium
View Solution$PCl_{5}$ dissociates as $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$. $5 \, \text{moles}$ of $PCl_{5}$ are placed in a $200 \, L$ vessel which contains $2 \, \text{moles}$ of $N_{2}$ and is maintained at $600 \, K$. The equilibrium pressure is $2.46 \, atm$. The equilibrium constant $K_{p}$ for the dissociation of $PCl_{5}$ is $...... \times 10^{-3}$. (nearest integer) (Given: $R = 0.082 \, L \, atm \, K^{-1} \, mol^{-1}$: Assume ideal gas behaviour)
Which of the following equations is incorrect?
Medium
View SolutionUsing the data provided,find the value of the equilibrium constant for the following reaction at $298 \ K$ and $1 \ atm$ pressure: $NO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{2(g)}$
$\Delta_{f} H^0(NO_{(g)}) = 90.4 \ kJ \cdot mol^{-1}$
$\Delta_{f} H^0(NO_{2(g)}) = 32.48 \ kJ \cdot mol^{-1}$
$\Delta S^{\circ} = -70.8 \ J \cdot K^{-1} \cdot mol^{-1}$
$\text{antilog}(6.4) = 2.51 \times 10^6$ (Note: Calculation based on standard thermodynamic relations)
$\Delta_{f} H^0(NO_{(g)}) = 90.4 \ kJ \cdot mol^{-1}$
$\Delta_{f} H^0(NO_{2(g)}) = 32.48 \ kJ \cdot mol^{-1}$
$\Delta S^{\circ} = -70.8 \ J \cdot K^{-1} \cdot mol^{-1}$
$\text{antilog}(6.4) = 2.51 \times 10^6$ (Note: Calculation based on standard thermodynamic relations)
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