Question

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

Answer 1

Answer:

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

$4 PH3(g)\rightarrow P4(g) + 6 H2(g)$

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

$Rate = +\frac{1}{6}*\frac{\Delta [H2]]}{\Delta t}$

$Rate = +\frac{1}{6}*\frac{C2[H2]-C1[H2]}{\Delta t}$

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

Δt = 93.0 s

$Rate = \frac{(0.101-0.0)M}{93.0 s} =1.09*10^{-3} M/s$

Since molarity M = mole/L

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