Exams › JEE Advanced › Chemistry

For the dissociation 2 NOBr(g) <=> 2 NO(g) + Br2(g), nitrosyl bromide is found to be (100/3)% dissociated at 25 deg C under a total pressure of 0.28 atm. Determine Kp at this temperature.

1. 1.61*10⁻³ atm
2. 2.34*10⁻³ atm
3. 3.22*10⁻³ atm
4. 6.44*10⁻³ atm

Correct answer: 1.61*10⁻³ atm

Solution

With alpha = 1/3, the equilibrium mole fractions give Kp = (2*alpha)²*(alpha) / [(2(1-alpha))²*(2+alpha)] * P_total, evaluating to about 1.61*10⁻³ atm.

Related JEE Advanced Chemistry questions

• In which of the following cases can the pH be determined using the equilibrium constant(s) provided?
• In the equilibrium reaction NH2COONH4(s) ⇌ 2NH3(g) + CO2(g), which change will cause the partial pressure of NH3 to rise?
• For the reaction XY2(g) ⇌ XY(g) + Y(g), if the value of α (degree of dissociation) is very small compared to 1, how is α related to the pressure (P)?
• At 460°C, 0.02 g of hydrogen reacts with 2.54 g of iodine until equilibrium is reached. Upon analysis, the equilibrium mixture contains 0.0021 moles of iodine. What is the equilibrium constant (Kc) for this reaction?
• Determine the pH of a 1.0 M solution of an unidentified hydroxide compound, assuming the metal ion has a +1 charge.
• What explains the acidic nature of an aqueous solution of aluminum chloride?

⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →