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3 years ago

Consider the reaction

Chemistry

2 answers:

spayn [35]3 years ago

7 0

Answer:

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.

Ratling [72]3 years ago

7 0

a. P₄ is changing at rate : 0.04 M/s

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

<h3>Further explanation </h3>

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 :

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

<h3>Learn more </h3>

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Keywords: reaction rate, molar concentration, products, reactants

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