Question

Be sure to answer all parts.

The thermal decomposition of phosphine \(\left(\mathrm{PH}_{3}\right)\) into phosphorus and molecular hydrogen is a first-order reaction:

\(4 \mathrm{PH}_{3}(g) \rightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g)\)

The half-life of the reaction is \(35.0 \mathrm{~s}\) at \(680^{\circ} \mathrm{C}\)

Calcnlate the first-order rate constant for the reaction:

_______s^-1

Calculate the time required for 95.0 percent of the phosphine to decompose:

_______Col

Answer 1

4 PH₃ lg) * PA (g) + 6H₂(gs. For first order reaction, ty = 0.693 Where tye - Half life K= rate constant. K = 0.693 tiz 35.0 ty amount K = 0.893 = 0.693 - 0.0198 sec ' [K2 0.01985-1] teten [PH3). [PH3).- initial, mercado) [PH3 ]- amount t= 0.0198 Xenos os -0.05 ao (as 95% reacted) * t = 151.3061 [PH₂ 12 after timet . In ao - 0.0198 - 151.30 s t=151.30 8