4PH 3 (g) P 4 (g) + 6H 2 (g) is the givenequation. d[P 4 ]/dt =4.5×10 -3 M/s at

Question

4PH3(g) P4(g) + 6H2(g) is the givenequation.
d[P4]/dt =4.5×10-3 M/s at a particular temperature and setof concentrations.

What are d[H2]/dt andd[PH3]/dt?

I see that as P4 is the given rate, as P4 makes 1 mol, H2 makes 6and PH3 4 mol.
I took the ratios of 6mol H2 / 1 mol P4 = 6 and 4 mol PH3 /1 =4

I then multiplied:
6*4.5e-3= 0.027 M/s
4*4.5e-3= 0.018 M/s

It then asks What is the rate of thereaction?

So wouldn't the rate of the reaction be the slowest rate of themall, that being the given?
4.5e-3 M/s ?

Thank You in Advance. I will rate you well for helping me withthis. I know it is obvious, the answer is just eluding me itseems.

Explanation / Answer

We Know that :         The given equationis:         4PH3(g) P4(g) + 6H2(g)       The rate of the reaction forthe above chemical reaction can be written as:          Rate of thereaction = (- 1/ 4 ) d [ PH3] / dt =   d[ P4 ] / dt = (1 / 6 ) d [H2 ] / dt             The rate of the d[P4]/dt =4.5×10-3 M/s so we have to calculate therate of PH3 and H2 as:             d [ PH3 ] / dt =   - 4x  4.5×10-3 M/s                                    = - 18 x 10-3 M/s             d [ H2 ] / dt = 6x  4.5×10-3 M/s                                  = 27 x 10-3 M/s           so therate of the reaction can be taken as the change in the rateof any one of the reactant or the product.

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---