Chapter Exam

Unit 6: Equilibrium — Online MCQ Test

Q: 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:

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.

Q: 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).

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.

Q: Which of the following mixture gives a buffer solution with pH 9.25? Given: pKb(NH4OH) = 4.75.

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.

Q: $25\ \mathrm{mL}$ 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

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 $\mathrm{I^-}$.

Q: 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:

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.

CHEMISTRY · JEE MAINS · JEE

Practice Unit 6: Equilibrium with a free chapter-wise online MCQ test for JEE JEE MAINS CHEMISTRY. 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.

Questions

20m

Time Limit

Attempts Left

10 random questions from this chapter (mixed difficulty)
Questions you've seen before won't repeat until the pool resets
You have 20 minutes — exam auto-submits when time is up
Maximum 3 attempts per chapter
Results and explanations shown immediately after submission

New here? Register free — includes 3 free chapter exams.

Unit 6: Equilibrium — Important Questions & Answers (FAQ)

Frequently asked questions from JEE JEE MAINS CHEMISTRY — Unit 6: Equilibrium, with answers and explanations. These are sample questions; the exam has its own separate question set.

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.

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.

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.

$25\ \mathrm{mL}$ 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 $\mathrm{I^-}$.

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.

← Previous Chapter

Unit 5: Solutions

Next Chapter →

Unit 7: Redox Reactions and Electrochemistry