Chemistry, 27.01.2020 22:31 jbrown76241

Consider the reaction

4ph3(g) → p4(g) + 6h2(g)

suppose that, at a particular moment during the reaction, molecular hydrogen is being formed at rate of 0.240 m/s.

(a) at what rate is p4 changing?

m/s

(b) at what rate is ph3 changing?

m/sf

Answers: 2

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Answers

Answer from: nataoli3843

1- 0.04 M/s.

2- 0.16 M/s.

Explanation:

For the reaction: 4PH₃ → P₄ + 6H₂.

The rate of the reaction = - d[PH₃]/4dt = d[P₄]/dt = d[H₂]/6dt.

where, - d[PH₃]/dt is the rate of PH₃ changing "rate of disappearance of PH₃".

d[P₄]/dt is rate of P₄ changing "rate of appearance of P₄".

d[H₂]/dt is the rate of H₂ changing "rate of formation of H₂" (d[H₂]/dt = 0.24 M/s).

(a) At what rate is P₄ changing?

∵ The rate of the reaction = d[P₄]/dt = d[H₂]/6dt.

∴ rate of P₄ changing = d[P₄]/dt = d[H₂]/6dt = (0.240 M/s)/(6.0) = 0.04 M/s.

(b) At what rate is PH₃ changing?

∵ The rate of the reaction = - d[PH₃]/4dt = d[H₂]/6dt.

∴ rate of PH₃ changing = - d[PH₃]/dt = 4(d[H₂]/6dt) = (4)(0.240 M/s)/(6.0) = 0.16 M/s.

Answer from: adammartinez12

a. P₄ is changing at rate : 0.04 M/s

b. PH₃ is changing at rate : -0.16 M/s

Further explanation

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.

Can be formulated:

Reaction: aA ---> bB

$\large{\boxed{\boxed{\bold{v~=~-\frac{\Delta A}{\Delta t}}}}$

$\large{\boxed{\boxed{\bold{v~=~+\frac{\Delta B}{\Delta t}}}}$

A = reagent

B = product

v = reaction rate

t = reaction time

For the reaction :

4PH₃(g) → P₄(g) + 6H₂(g)

Reaction rate is proportional with reaction coefficient :

$\displaystyle v=-\frac{1}{4}\frac{\Delta [PH_3]}{\Delta t}=+\frac{\Delta [P_4]}{\Delta t}=+\frac{1}{6}\frac{\Delta [H_2]}{\Delta t}$

Rate of H₂ = 0.24 M/s

then :

a. Rate of P₄ :

$\displaystyle v_{P_4}=\frac{1}{6}v_{H_2}\\\\v_{P_4}=\frac{1}{6}\times 0.24\\\\v_{P_4}=0.04~M/s$

b. Rate of PH₃ :

$\displaystyle v_{PH_3}=-\frac{4}{6}v_{H_2}\\\\v_{PH_3}=-\frac{4}{6}\times 0.24\\\\v_{P_4}=-0.16~M/s$

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Consider the reaction

4ph3(g) → p4(g) + 6h2(g)

suppose that, at a particular...

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