Consider the reaction shown below.
4 PH3(g) ==> P4(g) + 6 H2(g)
If, in a certain experiment, over a specific time period, 0.0035 mol PH3 is consumed in a 1.5 L container each second of reaction, what are the rates of production of P4 and H2 in this experiment?
P4 = ??? mol/Ls
H2 = ??? mol/Ls
2007-04-02 06:05:23 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
r(PH3)=(0.0035/1.5)/1=0.0035 mol.L-1.s-1
r(PH3)=4r(P4)=(6/4)r(H2)
r(P4)=0,0035/4=8.75x10^-4 mol.L-1.s-1
r(H2)=1.5x0.0035=0.0054 mol.L-1.s-1
GOOD LUCK!
2007-04-10 03:01:25 · answer #1 · answered by Master of Chemistry 2 · 0⤊ 0⤋
P4 appears at 1/4 of the rate that PH3 disappears, and H2 appears at 1.5 x the rate at which PH3 disappears.
2007-04-02 06:35:20 · answer #2 · answered by Gervald F 7 · 0⤊ 0⤋
P4= 0.0035/ (1.5*4) mol/Ls
= 0.00055333 mol/Ls
H2= 6*0.0035/ (1.5*4) mol/Ls
= 0.0035 mol/Ls
2007-04-09 20:44:49 · answer #3 · answered by 99%cocoa 1 · 0⤊ 0⤋