Applications Of Equilibrium Constants

Step 1: Formation of complex reduces free drastically due to excess CN .
Step 2: Using
Step 3: Substituting given concentrations and solving gives

Step 1: Equilibrium constant: Kp = PB² PC³
Step 2: If PC → 2PC: PB² ∝ (1)/((2)³) ⟹ PB ∝ \frac12√(2)

Step 1: For the reaction, total gaseous moles =3.
Step 2: Δ G^∘ = -RTln Kp
Step 3: Since partial pressures are proportional to X, Kp ∝ X³ ⟹ Δ G^∘ = -3RTln X

Step 1: Equilibrium constant: Kp = PB² PC³
Step 2: If PC is doubled: PC → 2PC
Step 3: To keep Kp constant: PB²(2PC)³=PB'²PC³ ⟹ PB' = 2√(2)PB

Step 1: {Understanding the Problem}
The given reaction is the combustion of benzene. We are given the heat of combustion at constant volume, and we need to find the heat of combustion at constant pressure. For the combustion of benzene: Step 2: {Using the Relation between Heat at Constant Volume and Pressure}
The change in enthalpy is related to the change in internal energy by the equation: where is the change in the number of moles of gas between products and reactants. In the given reaction, we calculate the change in the number of moles of gas: Thus, we have: Thus, the correct answer is (D).