1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
fredd [130]
2 years ago
6

The decomposition reaction of N2O5 in carbon tetrachloride is 2N2O5−→−4NO2+O2. The rate law is first order in N2O5. At 64 °C the

rate constant is 4.82×10−3s−1. (a) Write the rate law for the reaction. (b) What is the rate of reaction when [N2O5]=0.0240M? (c) What happens to the rate when the concentration of N2O5 is doubled to 0.0480 M? (d) What happens to the rate when the concentration of N2O5 is halved to 0.0120 M?
Chemistry
2 answers:
Sophie [7]2 years ago
7 0

Answer:

a) r = 4.82x10⁻³*[N2O5]

b) 1.16x10⁻⁴ M/s

c) The rate is doubled too (2.32x10⁻⁴ M/s)

d) The rate is halved too (5.78x10⁻⁴ M/s)

Explanation:

a) The rate law of a generic reaction (A + B → C + D) can be expressed by:

r = k*[A]ᵃ*[B]ᵇ

Where k is the rate constant, [X] is the concentration of the compound X, and a and b are the coefficients of the reaction (which can be different from the ones of the chemical equation).

In this case, there is only one reactant, and the reaction is first order, which means that a = 1. So, the rate law is:

r = k*[N2O5]

r = 4.82x10⁻³*[N2O5]

b) Substituing the value of the concentration in the rate law:

r = 4.82x10⁻³*0.0240

r = 1.16x10⁻⁴ M/s

c) When [N2O5] = 0.0480 M,

r = 4.82x10⁻³*0.0480

r = 2.32x10⁻⁴ M/s

So, the rate is doubled too.

d) When [N2O5] = 0.0120 M,

r = 4.82x10⁻³*0.0120

r = 5.78x10⁻⁴ M/s

So, the rate is halved too.

Mademuasel [1]2 years ago
3 0

Answer:

(a) rate =  4.82 x 10⁻³s⁻¹  [N2O5]

(b) rate =   1.16 x 10⁻⁴  M/s

(c) rate =   2.32 x 10⁻⁴ M/s

(d) rate =   5.80 x 10⁻⁵ M/s

Explanation:  

We are told the rate law is first order in N₂O₅, and its rate constant is 4.82 x 10⁻³s⁻¹ . This means the rate is proportional to the molar concentration   of   N₂O₅, so

(a) rate = k [N2O5] = 4.82 x 10⁻³s⁻¹ x [N2O5]

(b) rate = 4.82×10⁻³s⁻¹  x 0.0240 M =  1.16 x 10⁻⁴ M/s

(c) Since the reaction is first order if the concentration of  N₂O₅ is double the rate will double too:  2 x   1.16 x 10⁻⁴ M/s = 2.32 x 10⁻⁴ M/s

(d) Again since the reaction is halved to 0.0120 M, the rate will be halved to

1.16 x 10⁻⁴ M/s / 2 =  5.80 x 10⁻⁵ M/s

You might be interested in
What two quantities must be known to calculate the density of a sample of matter? A) mass and volume B) length and mass C) size
Svetradugi [14.3K]

Answer:

A

Explanation:

4 0
2 years ago
Read 2 more answers
1
Alex787 [66]

Answer:

The reaction of an acid and a base is called a neutralization reaction. ... However, in the reaction between HCl(aq) and Mg(OH) 2(aq), additional molecules of HCl and ... First, we will write the chemical equation with the formulas of the reactants ... The chloride ions are the only spectator ions  I THINKS

Explanation:

4 0
3 years ago
I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol
galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = \Delta H_f_{(I_2)}=62.438 kJ/mol

Enthalpy of formation of chlorine gas = \Delta H_f_{(Cl_2)}=0 kJ/mol

Enthalpy of formation of ICl gas = \Delta H_f_{(ICl)}=17.78 kJ/mol

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]

=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol

Enthaply change when 1.62 moles of iodine gas recast:

\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ

Entropy of the surrounding = \Delta S^o_{surr}=\frac{\Delta H}{T}

=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

4 0
3 years ago
Calculate the morality of a 35.4% (by mass) aqueous solution of phosphoric acid (H3PO4). Molar mass of H3PO4 is 98.00g/mol
Butoxors [25]

Answer:

no idea with the answer pls check with otherr

3 0
2 years ago
What does mass change to and what does it do
Nataly [62]

Answer:

Mass is the amount of matter in an object and does not change with location.

Explanation:

4 0
3 years ago
Other questions:
  • If a country wants to increase its GPA what actions could be taken
    7·1 answer
  • The formula for a compound of Li+ ions and Br- ions is written LiBr. Why can’t it be written Li2Br? Why isn’t it written BrLi?
    5·1 answer
  • Emission nebulae are also called ____ because they are composed of ionized hydrogen.
    11·1 answer
  • In group 14 which is a metalloid
    10·1 answer
  • What is the name given to the property of metal that allows electricity to pass through.
    10·2 answers
  • An éclair is made from which of the following types of dough?
    8·2 answers
  • HELP PLEASE :((( Explain why helium filled balloons deflate over time faster than air filled balloons do?
    15·2 answers
  • How does organic matter help soil?
    14·2 answers
  • How many grams are present in 0.885 moles of manganese?
    10·1 answer
  • Please hurry and tell the answer to:
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!