Top 50 Most Repeated EQUILIBRIUM PYQs | JEE MAINS
A curated collection of the most important questions from EQUILIBRIUM, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from EQUILIBRIUM, fully solved with step-by-step concepts to prepare for JEE MAINS.
$K_p$ = (partial pressures of products)^stoich / (reactants)^stoich...
Read Full Step-by-Step Solution →Using the Henderson–Hasselbalch equation: $\text{pH}=\text{p}K_a+\log\frac{[A^-]}{[HA]} = 4.76+\log\frac{0.10}{0.20}=4.76+\log0.5 = 4.76-0.301 = 4.459...
Read Full Step-by-Step Solution →The relationship is $K_p = K_c (RT)^{\Delta n}$, where $\Delta n = 2 - (1 + 3) = -2$. So $K_p = K_c (RT)^{-2}$....
Read Full Step-by-Step Solution →$\Delta n = -1$, so $K_p = K_c(RT)^{-1} = 4/(0.0821\times300) \approx 0.16$...
Read Full Step-by-Step Solution →With the common ion \(\ce{Cl^-}\) from NaCl, \([\ce{Cl^-}] \approx 0.010\ \text{M}\). Using \(K_{sp}= [\ce{Ag^+}][\ce{Cl^-}]\), we get \([\ce{Ag^+}] =...
Read Full Step-by-Step Solution →Foundational check for Equilibrium. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →The equilibrium constant expression for the reaction is Kp = (PCO2) / (PCO). Using the given Kp value and initial pressure of CO, we can calculate the...
Read Full Step-by-Step Solution →$K_p$ = (partial pressures of products)^stoich / (reactants)^stoich...
Read Full Step-by-Step Solution →Using the Henderson–Hasselbalch equation: $\text{pH}=\text{p}K_a+\log\frac{[A^-]}{[HA]} = 4.76+\log\frac{0.10}{0.20}=4.76+\log0.5 = 4.76-0.301 = 4.459...
Read Full Step-by-Step Solution →The relationship is $K_p = K_c (RT)^{\Delta n}$, where $\Delta n = 2 - (1 + 3) = -2$. So $K_p = K_c (RT)^{-2}$....
Read Full Step-by-Step Solution →$\Delta n = -1$, so $K_p = K_c(RT)^{-1} = 4/(0.0821\times300) \approx 0.16$...
Read Full Step-by-Step Solution →With the common ion \(\ce{Cl^-}\) from NaCl, \([\ce{Cl^-}] \approx 0.010\ \text{M}\). Using \(K_{sp}= [\ce{Ag^+}][\ce{Cl^-}]\), we get \([\ce{Ag^+}] =...
Read Full Step-by-Step Solution →Foundational check for Equilibrium. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →The equilibrium constant expression for the reaction is Kp = (PCO2) / (PCO). Using the given Kp value and initial pressure of CO, we can calculate the...
Read Full Step-by-Step Solution →$K_p$ = (partial pressures of products)^stoich / (reactants)^stoich...
Read Full Step-by-Step Solution →Using the Henderson–Hasselbalch equation: $\text{pH}=\text{p}K_a+\log\frac{[A^-]}{[HA]} = 4.76+\log\frac{0.10}{0.20}=4.76+\log0.5 = 4.76-0.301 = 4.459...
Read Full Step-by-Step Solution →The relationship is $K_p = K_c (RT)^{\Delta n}$, where $\Delta n = 2 - (1 + 3) = -2$. So $K_p = K_c (RT)^{-2}$....
Read Full Step-by-Step Solution →$\Delta n = -1$, so $K_p = K_c(RT)^{-1} = 4/(0.0821\times300) \approx 0.16$...
Read Full Step-by-Step Solution →With the common ion \(\ce{Cl^-}\) from NaCl, \([\ce{Cl^-}] \approx 0.010\ \text{M}\). Using \(K_{sp}= [\ce{Ag^+}][\ce{Cl^-}]\), we get \([\ce{Ag^+}] =...
Read Full Step-by-Step Solution →Foundational check for Equilibrium. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →The equilibrium constant expression for the reaction is Kp = (PCO2) / (PCO). Using the given Kp value and initial pressure of CO, we can calculate the...
Read Full Step-by-Step Solution →$K_p$ = (partial pressures of products)^stoich / (reactants)^stoich...
Read Full Step-by-Step Solution →Using the Henderson–Hasselbalch equation: $\text{pH}=\text{p}K_a+\log\frac{[A^-]}{[HA]} = 4.76+\log\frac{0.10}{0.20}=4.76+\log0.5 = 4.76-0.301 = 4.459...
Read Full Step-by-Step Solution →The relationship is $K_p = K_c (RT)^{\Delta n}$, where $\Delta n = 2 - (1 + 3) = -2$. So $K_p = K_c (RT)^{-2}$....
Read Full Step-by-Step Solution →$\Delta n = -1$, so $K_p = K_c(RT)^{-1} = 4/(0.0821\times300) \approx 0.16$...
Read Full Step-by-Step Solution →With the common ion \(\ce{Cl^-}\) from NaCl, \([\ce{Cl^-}] \approx 0.010\ \text{M}\). Using \(K_{sp}= [\ce{Ag^+}][\ce{Cl^-}]\), we get \([\ce{Ag^+}] =...
Read Full Step-by-Step Solution →Foundational check for Equilibrium. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →The equilibrium constant expression for the reaction is Kp = (PCO2) / (PCO). Using the given Kp value and initial pressure of CO, we can calculate the...
Read Full Step-by-Step Solution →$K_p$ = (partial pressures of products)^stoich / (reactants)^stoich...
Read Full Step-by-Step Solution →Using the Henderson–Hasselbalch equation: $\text{pH}=\text{p}K_a+\log\frac{[A^-]}{[HA]} = 4.76+\log\frac{0.10}{0.20}=4.76+\log0.5 = 4.76-0.301 = 4.459...
Read Full Step-by-Step Solution →The relationship is $K_p = K_c (RT)^{\Delta n}$, where $\Delta n = 2 - (1 + 3) = -2$. So $K_p = K_c (RT)^{-2}$....
Read Full Step-by-Step Solution →$\Delta n = -1$, so $K_p = K_c(RT)^{-1} = 4/(0.0821\times300) \approx 0.16$...
Read Full Step-by-Step Solution →With the common ion \(\ce{Cl^-}\) from NaCl, \([\ce{Cl^-}] \approx 0.010\ \text{M}\). Using \(K_{sp}= [\ce{Ag^+}][\ce{Cl^-}]\), we get \([\ce{Ag^+}] =...
Read Full Step-by-Step Solution →Foundational check for Equilibrium. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →The equilibrium constant expression for the reaction is Kp = (PCO2) / (PCO). Using the given Kp value and initial pressure of CO, we can calculate the...
Read Full Step-by-Step Solution →$K_p$ = (partial pressures of products)^stoich / (reactants)^stoich...
Read Full Step-by-Step Solution →Using the Henderson–Hasselbalch equation: $\text{pH}=\text{p}K_a+\log\frac{[A^-]}{[HA]} = 4.76+\log\frac{0.10}{0.20}=4.76+\log0.5 = 4.76-0.301 = 4.459...
Read Full Step-by-Step Solution →The relationship is $K_p = K_c (RT)^{\Delta n}$, where $\Delta n = 2 - (1 + 3) = -2$. So $K_p = K_c (RT)^{-2}$....
Read Full Step-by-Step Solution →$\Delta n = -1$, so $K_p = K_c(RT)^{-1} = 4/(0.0821\times300) \approx 0.16$...
Read Full Step-by-Step Solution →With the common ion \(\ce{Cl^-}\) from NaCl, \([\ce{Cl^-}] \approx 0.010\ \text{M}\). Using \(K_{sp}= [\ce{Ag^+}][\ce{Cl^-}]\), we get \([\ce{Ag^+}] =...
Read Full Step-by-Step Solution →Foundational check for Equilibrium. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →The equilibrium constant expression for the reaction is Kp = (PCO2) / (PCO). Using the given Kp value and initial pressure of CO, we can calculate the...
Read Full Step-by-Step Solution →$K_p$ = (partial pressures of products)^stoich / (reactants)^stoich...
Read Full Step-by-Step Solution →Using the Henderson–Hasselbalch equation: $\text{pH}=\text{p}K_a+\log\frac{[A^-]}{[HA]} = 4.76+\log\frac{0.10}{0.20}=4.76+\log0.5 = 4.76-0.301 = 4.459...
Read Full Step-by-Step Solution →