The first-order decomposition of phosphene to phosphorus and hydrogen, shown below, 4PH₃(g) $\rarr$ P₄(g) + 6H₂(g)
Has a half-life of 35.0 s at 680^$\circ$C. Starting with 520 mmHg of pure phosphene in an 8.00-L flask at 680^$\circ$C, how long will it take for the total pressure in the flask to rise to 1.000 atm

Question

The first-order decomposition of phosphene to phosphorus and hydrogen, shown below, 4PH₃(g)

\rarr

P₄(g) + 6H₂(g)
Has a half-life of 35.0 s at 680^$\circ$C. Starting with 520 mmHg of pure phosphene in an 8.00-L flask at 680^$\circ$C, how long will it take for the total pressure in the flask to rise to 1.000 atm

Quizplus · Accepted Answer

The decomposition of phosphene is a first-order reaction, and the half-life is given as 35.0 s. The initial pressure of phosphene is 520 mmHg, which is equivalent to 0.684 atm. The total pressure will rise to 1.000 atm when the reaction is complete. Using the first-order kinetics equation and the given half-life, the time required for the pressure to reach 1.000 atm is approximately 48.3 s.

The First-Order Decomposition of Phosphene to Phosphorus and Hydrogen, Shown →\rarr→

The First-Order Decomposition of Phosphene to Phosphorus and Hydrogen, Shown $\rarr$