## Chemistry and Chemical Reactivity (9th Edition)

a) From the rate law $r=k[PH_3]$, we deduce it's a first-order reaction, for which: $\ln(C/C_0)=-0.693/t_{1/2}\times t$ Decomposing 3/4 means that $C/C_0=0.25$ $\ln 0.25=-0.693/37.9\ s\times t$ $t=75.8\ s$ b) $\ln (C/C_0) = -0.693\times 60\ s/37.9\ s$ $(C/C_0)=0.334\ or\ 33.4\%$ The remaining fraction is 33.4% of the original.