Question

The decomposition of PH3 proceeds according to the following equation

4PH3(g)-----------------> P4(g) + 6H2(g)

rate=k[PH3] , t1/2=37.9 s at 120 degrees

how much time is required for 3/4th of PH3 to decompose?

What is the concentration of PH3 after 1 min?

Answer 1

Answer:

Explanation:

T50% = 37.9sec.

K=?

T50%=0.693/k

K=0.693/37.9=(0.0182)

1. T=? For 3/4th of the ph3 to decompose

T=2.303/k log a/a-x

Here k=0.0182

a=1 x=3/4

T=2.303/0.0182 log 1/1-3/4

T=126.53 log 4

Log4=0.6021

T=126.53×0.6021

T=76.1 sec.

2. After 1 minutes that means

T=1 mint.

K=0.0182

a=1 x=?

0.0182 =2.303/1 log 1/1-x

0.0182 /2.303 =log 1/1-x

Antilog 0.00790=1/1-x

X=0.99