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Ugo [173]
3 years ago
9

The Haber reaction for the manufacture of ammonia is: N2 + 3H2 → 2NH3 Without doing any experiments, which of the following can

you say MUST be true? Disappearance rate of H2 = 3 (Disappearance rate of N2). The reaction is first order in N2. Reaction rate = -Δ[N2]/Δt. The reaction is not an elementary reaction. Δ[H2]/Δt will have a positive value. Disappearance rate of N2 = 3 (Disappearance rate of H2). The activation energy is positive.
Chemistry
1 answer:
Dimas [21]3 years ago
3 0

Answer :  The correct statement is, \text{Rate of disappearance of }H_2=3\times (\text{Rate of disappearance of }N_2)

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,

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

The expression for rate of reaction :

\text{Rate of disappearance of }N_2=-\frac{d[N_2]}{dt}

\text{Rate of disappearance of }H_2=-\frac{1}{3}\frac{d[H_2]}{dt}

\text{Rate of formation of }NH_3=+\frac{1}{2}\frac{d[NH_3]}{dt}

From this we conclude that,

\text{Rate of disappearance of }H_2=3\times (\text{Rate of disappearance of }N_2)

Hence, the correct statement is, \text{Rate of disappearance of }H_2=3\times (\text{Rate of disappearance of }N_2)

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8 0
4 years ago
An organic compound of relative molecular mass 46,on analysis was found to contain 52.0% carbon, 13.3% hydrogen and 34.7% oxygen
Nadya [2.5K]

Answer:

Empirical formula is C₂H₆O.

Explanation:

Empirical formula:

It is the simplest formula gives the ratio of atoms of different elements in small whole number

Given data:

Percentage of hydrogen = 13.3%

Percentage of carbon = 52.0%

Percentage of oxygen = 34.7%

Empirical formula = ?

Molecular formula = ?

Solution:

Number of gram atoms of H = 13.3 / 1.01 = 13 .17

Number of gram atoms of O = 34.7 / 16 = 2.17

Number of gram atoms of C = 52.0 / 12 = 4.3

Atomic ratio:

            C                     :          H             :           O

           4.3/2.17            :     13.17/2.17     :       2.17/2.17

            2                      :        6               :        1

C : H : O = 2 : 6 : 1

Empirical formula is C₂H₆O.

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

n = 46 / 158

The relative molecular mass of compound is not correct.

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Balance the following equation and answer the question. ___CuCl + ___ H2S ------ ___ Cu2S + ___ HCl If 1.2 moles of CuCl is mixe
brilliants [131]

Answer:

A. 2CuCl + H2S —> Cu2S + 2HCl

B. 0.6 mole of Cu2S

Explanation:

A. The balancing equation.

The equation can be balance as follow:

CuCl + H2S —> Cu2S + HCl

There are 2 atoms of Cu on the right side and 1 atom on the left. It can be balance by putting 2 in front of CuCl as shown below:

2CuCl + H2S —> Cu2S + HCl

There are 2 atoms of H on the left side and 1 atom on the right side. It can be balance by putting 2 in front of HCl as shown below:

2CuCl + H2S —> Cu2S + 2HCl

Now we can see that the equation is balanced.

B. Determination of the number of mole of Cu2S produced by the reaction 1.2 mole of CuCl.

This is illustrated below:

2CuCl + H2S —> Cu2S + 2HCl

From the balanced equation above, 2 moles of CuCl reacted to produce 1 mole of Cu2S.

Therefore, 1.2 mole of CuCl will react to produce = 1.2/2 = 0.6 mole of Cu2S.

Therefore, 0.6 mole of Cu2S is produced from the reaction of 1.2 mole of CuCl.

3 0
4 years ago
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