Unit 6: Equilibrium — Online MCQ Test
CHEMISTRY · JEE MAINS · JEE
Practice Unit 6: Equilibrium with a free chapter-wise online MCQ test.
This chapter covers: Meaning of equilibrium is the concept of dynamic equilibrium. Equilibria involving physical processes: Solid-liquid liquid-gas gas-gas and solid-gas equilibria Henry's law. General....
AI-generated questions from basic to board-exam level, with instant results and explanations.
Unit 6: Equilibrium — Important Questions & Answers
A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of CO2 is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm, then Kp is:
- A. 1.8 atm
- B. 0.3 atm
- C. 3 atm
- D. 0.18 atm
Answer: A. 1.8 atm
CO2 + C(s) ⇌ 2CO(g). Initial: p(CO2) = 0.5, p(CO) = 0. Equil: p(CO2) = 0.5-x, p(CO) = 2x. Total pressure = (0.5-x) + 2x = 0.5+x = 0.8 => x=0.3. So p(CO2)=0.2, p(CO)=0.6. Kp = (0.6^2)/0.2 = 0.36/0.2 = 1.8. Wait, checking calculation: Kp = 1.8 is option 1, but prompt says answer key is 3.
CO2 + C(s) ⇌ 2CO(g). Initial: p(CO2) = 0.5, p(CO) = 0. Equil: p(CO2) = 0.5-x, p(CO) = 2x. Total pressure = (0.5-x) + 2x = 0.5+x = 0.8 => x=0.3. So p(CO2)=0.2, p(CO)=0.6. Kp = (0.6^2)/0.2 = 0.36/0.2 = 1.8. Wait, checking calculation: Kp = 1.8 is option 1, but prompt says answer key is 3.
The first and second ionization constants of H2X are 2.5 × 10^-8 and 1.0 × 10^-13 respectively. The concentration of X^2- in 0.1M H2X solution is x × 10^-15 M. Find the value of x (Nearest Integer).
- A. 100
- B. 50
- C. 10
- D. 1
Answer: A. 100
For a diprotic acid where K1 >> K2, the concentration of the dianion [X^2-] is approximately equal to the second dissociation constant K2. Here K2 = 1.0 × 10^-13 = 100 × 10^-15, so x = 100.
For a diprotic acid where K1 >> K2, the concentration of the dianion [X^2-] is approximately equal to the second dissociation constant K2. Here K2 = 1.0 × 10^-13 = 100 × 10^-15, so x = 100.
Which of the following mixture gives a buffer solution with pH 9.25? Given: pKb(NH4OH) = 4.75.
- A. 0.2M NH4OH 0.4L + 0.1M HCl 1L
- B. 0.2M NH4OH 0.5L + 0.1M HCl 0.5L
- C. 0.5M NH4OH 0.2L + 0.2M HCl 0.5L
- D. 0.4M NH4OH 1L + 0.1M HCl 1L
Answer: B. 0.2M NH4OH 0.5L + 0.1M HCl 0.5L
For a basic buffer, pOH = pKb + log([Salt]/[Base]). For pH = 9.25, pOH = 4.75. Since pOH = pKb, we need log([Salt]/[Base]) = 0, so [Salt] = [Base]. Option (2) gives 0.05 moles NH4Cl (salt) and 0.05 moles NH4OH (base remaining: 0.1-0.05 = 0.05), satisfying the buffer condition.
For a basic buffer, pOH = pKb + log([Salt]/[Base]). For pH = 9.25, pOH = 4.75. Since pOH = pKb, we need log([Salt]/[Base]) = 0, so [Salt] = [Base]. Option (2) gives 0.05 moles NH4Cl (salt) and 0.05 moles NH4OH (base remaining: 0.1-0.05 = 0.05), satisfying the buffer condition.
$25\ \mathrm{mL}$ of silver nitrate solution $(1\ \mathrm{M})$ is added dropwise to $25\ \mathrm{mL}$ of potassium iodide $(1.05\ \mathrm{M})$ solution. The ion(s) present in very small quantity in the solution is/are
- A. $\mathrm{I^-}$ only
- B. $\mathrm{Ag^+}$ and $\mathrm{I^-}$ both
- C. $\mathrm{K^+}$ only
- D. $\mathrm{NO_3^-}$ only
Answer: B. $\mathrm{Ag^+}$ and $\mathrm{I^-}$ both
Moles of $\mathrm{Ag^+}$ added $= 0.025$ mol. Moles of $\mathrm{I^-}$ present $= 0.025\times 1.05 = 0.02625$ mol. Since $\mathrm{AgI}$ precipitates in a $1:1$ ratio, all $\mathrm{Ag^+}$ is consumed and excess $\mathrm{I^-}$ remains. However, because the salt is sparingly soluble, the equilibrium concentration of both $\mathrm{Ag^+}$ and $\mathrm{I^-}$ in solution is very small, while $\mathrm{K^+}$ and $\mathrm{NO_3^-}$ remain as spectator ions in much larger concentration. Therefore the ions present in very small quantity are both $\mathrm{Ag^+}$ and $\mathrm{I^-}$.
Moles of $\mathrm{Ag^+}$ added $= 0.025$ mol. Moles of $\mathrm{I^-}$ present $= 0.025\times 1.05 = 0.02625$ mol. Since $\mathrm{AgI}$ precipitates in a $1:1$ ratio, all $\mathrm{Ag^+}$ is consumed and excess $\mathrm{I^-}$ remains. However, because the salt is sparingly soluble, the equilibrium concentration of both $\mathrm{Ag^+}$ and $\mathrm{I^-}$ in solution is very small, while $\mathrm{K^+}$ and $\mathrm{NO_3^-}$ remain as spectator ions in much larger concentration. Therefore the ions present in very small quantity are both $\mathrm{Ag^+}$ and $\mathrm{I^-}$.
0.01 M जलीय $\mathrm{Ba(OH)_2}$ के 10.0 mL को उदासीन करने के लिए आवश्यक 0.02 M जलीय HBr का आयतन है:
- A. 5.0 mL
- B. 7.5 mL
- C. 10.0 mL
- D. 2.5 mL
Answer: C. 10.0 mL
$\mathrm{Ba(OH)_2}$ gives 2 $\mathrm{OH^-}$ per mole, so moles of base in 10.0 mL are $0.01\times 0.010 = 1.0\times 10^{-4}$ mol. Equivalent moles of $\mathrm{OH^-}$ are $2.0\times 10^{-4}$ mol, hence required moles of HBr are $2.0\times 10^{-4}$ mol. At 0.02 M, volume needed is $\dfrac{2.0\times 10^{-4}}{0.02} = 1.0\times 10^{-2}$ L $= 10.0$ mL.
$\mathrm{Ba(OH)_2}$ gives 2 $\mathrm{OH^-}$ per mole, so moles of base in 10.0 mL are $0.01\times 0.010 = 1.0\times 10^{-4}$ mol. Equivalent moles of $\mathrm{OH^-}$ are $2.0\times 10^{-4}$ mol, hence required moles of HBr are $2.0\times 10^{-4}$ mol. At 0.02 M, volume needed is $\dfrac{2.0\times 10^{-4}}{0.02} = 1.0\times 10^{-2}$ L $= 10.0$ mL.