For the gas phase decomposition of phosphine at 120 °c, the rate of the reaction is determined by measuring the appearance of h2. 4 ph3(g)p4(g) + 6 h2(g) at the beginning of the reaction, the concentration of h2 is 0 m. after 93.0 s the concentration has increased to 0.101 m. what is the rate of the reaction? (mol h2/l) /s

Rate = 1.09*10^-3 (mol H2/L)/s

Explanation:

Given:

Initial concentration of H2, C1 = 0 M

Final concentration of H2, C2 = 0.101 M

Time taken, t = 93.0 s

To determine:

The rate of the given reaction

Calculation:

The decomposition of PH3 is represented by the following chemical reaction

Reaction rate in terms of the appearance of H2 is given as:

Here C1(H2) = 0 M and C2(H2) = 0.101 M

Δt = 93.0 s

Since molarity M = mole/L

rate = 1.09*10^-3 (mol H2/L)/s

the answer is the

43.452

step-by-step explanation:

284 times .153 = 43.452