Question

At high temperatures phosphine (PH_3) dissociates into phosphorus and hydrogen by the following reaction: 4PH3 rightarrow P_4 + 6H_2 At 800 degree C the rate at which phosphine dissociates is dC_PH_3/dt = -3.715 times 10^-6 C_PH_3. for t in seconds. The reaction occurs in a constant-volume, 2-L vessel, and the initial concentration of phosphine is 5kmol/m^3

a. If 3mol of the phosphine reacts, how much phosphorus and hydrogen is produced?
b. Develop expressions for the number of moles of phosphine, phosphorus, and hydrogen present at any time, and determine how long it would take for 3 mol of phosphine to have reacted.

Answer 1

Answer:a)p4=0.75mol

H2=4.5mol

B)PH3=1×10^7mole

P4=0.25×10^7mole

H2=1.5×10^7mole

Time taken for 3mol of phosphine to react is 54 ×10^7s

Explanation:From the balanced chemical equation

4PH3____P4+6H2

From Gay lussac law ,when gases combines

,They do so in a ratio which bear a simple ratio with their products,if gases.

So it means from the equation above

4 mol of PH3 will form 1mol of P4 and 6 mol of H2

Meaning that if 4mol of PH3 gives 1mol of P4

Then 3mol of PH3 will give 1/4 mol or 0.75 mol of P4 .

In the same vein,3mol of PH3 will give 6×3/4 mol of H2= 4.5 Mol of H2

B) molar concentration =mole/volume

Meaning mole=molar concentration ×volume

Molar concentration is given as 5Km/m^3=5000mol/m^3

Volume=2L =2×1000=2000m^3

Nos of moles=5000×2000=10^7moles

So for P4,nos of moles =1/4×10^7=0.25×10^7

ForH2 ,6/4 ×10^7=1.5×10^7 moles

Rate= volume/time

Time taken for reaction will be volume/rate=2000/3.715×10^6

=54×10^7s