Final answer:
The partial pressure of nitric oxide (NO) at equilibrium, P(NO)ᴇᴢᴛᴛ, calculated using the provided equilibrium constant (Kp) and given partial pressures for NOCI and Cl₂, is 0.23 atm.
Step-by-step explanation:
To calculate the partial pressure of NO at equilibrium, we can apply the equilibrium constant (Kp) to the reaction 2 NOCI (g) ⇌ 2 NO (g) + Cl₂ (g). According to the question, P(NOCI)ᴇᴢᴛᴛ = 0.35 atm, P(Cl₂)ᴇᴢᴛᴛ = 0.40 atm, and Kp = 1.9 × 10⁻². Based on the stoichiometry of the reaction, the formation of Cl₂ corresponds to the formation of 2 moles of NO for every 1 mole of Cl₂ that forms. Because the pressures are initially zero for NO and given for Cl₂ at equilibrium, we can use the expression for Kp to find P(NO).
The expression for Kp is:
Kp = ( P(NO)² × P(Cl₂) ) / ( P(NOCI)² )
Let's insert the given information into this equation.
1.9 × 10⁻² = ( P(NO)² × 0.40 atm ) / ( 0.35 atm )²
We can then solve for P(NO)²:
P(NO)² = (1.9 × 10⁻²) × (0.35 atm)² / 0.40 atm
After performing the calculation, P(NO) = 0.23 atm (which matches option A from the multiple choices provided).
Therefore, the partial pressure of NO at equilibrium, P(NO)ᴇᴢᴛᴛ, is 0.23 atm.