Chemistry, 28.02.2020 19:52 sissygirl0807

When the rate of disappearance of O2 is 0.42 Ms−1, the rate of disappearance of C2H4 is Ms−1. When the rate of disappearance of is 0.42 , the rate of disappearance of is . 0.84 0.28 0.14 0.63 1.3

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Answer from: gabrielduboy

This is an incomplete question, here is a complete question.

The combustion of ethylene proceeds by the reaction

$C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(g)$

When the rate of disappearance of O2 is 0.42 Ms−1, the rate of disappearance of C₂H₄ is ________ Ms−1.

(a) 0.84 (b) 0.28 (c) 0.14 (d) 0.63 (e) 1.3

Answer : The rate of disappearance of C₂H₄ is, (c) 0.14 M/s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }C_2H_4=-\frac{d[C_2H_4]}{dt}$

$\text{Rate of disappearance of }O_2=-\frac{1}{3}\frac{d[O_2]}{dt}$

$\text{Rate of formation of }CO_2=+\frac{1}{2}\frac{d[CO_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{2}\frac{d[H_2O]}{dt}$

As we are given:

$\text{Rate of disappearance of }O_2=0.42M/s$

$-\frac{d[C_2H_4]}{dt}=-\frac{1}{3}\frac{d[O_2]}{dt}$

$\frac{d[C_2H_4]}{dt}=\frac{1}{3}\frac{d[O_2]}{dt}$

$\frac{d[C_2H_4]}{dt}=\frac{1}{3}\times 0.42M/s$

$\frac{d[C_2H_4]}{dt}=0.14M/s$

Thus, the rate of disappearance of C₂H₄ is, 0.14 M/s

Answer from: Quest

Read the sentences below in 1962, i tried to enroll as the first black student at the university of mississippi. the governor of mississippi blocked my entrance which best describes the verbs used in the sentences? o they are past tense verbs, which are often used in narrative essays o they are past tense verbs, which are not often used in narrative essays. o they are future tense verbs, which are often used in narrative essays o they are future tense verbs, which are not often used in narrative essays.

Answer from: Quest

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2

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When the rate of disappearance of O2 is 0.42 Ms−1, the rate of disappearance of C2H4 is Ms−1. When...