Answer:
e) The activation energy of the reverse reaction is greater than that of the forward reaction.
Explanation:
- Activation energy is the minimum amount of energy that is required by the reactants to start a reaction.
- An exothermic reaction is a reaction that releases heat energy to the surrounding while an endothermic reactions is a reaction that absorbs heat from the surrounding.
- <em><u>In reversible reactions, when the forward reaction is exothermic it means the reverse reaction will be endothermic, therefore the reverse reaction will have a higher activation energy than the forward reaction.</u></em> The activation energy of the reverse reaction will be the sum of the enthalpy and the activation energy of the forward reaction.
Answer:
3.64g
Explanation:
Given parameters:
Mass of NH₃ = 18.1g
Mass of Cu₂O = 90.4g
Unknown:
Limiting reactant = ?
Mass of N₂ formed = ?
Solution:
The reaction equation is given as:
Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O
The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;
Number of moles = 
Molar mass of Cu₂O = 2(63.6) + 16 = 143.2g/mol
Molar mass of NH₃ = 14 + 3(1) = 17g/mol
Number of moles of Cu₂O = 
= 0.13moles
Number of moles of NH₃ = 
= 5.32moles
From this reaction;
1 mole of Cu₂O combines with 2 mole of NH₃
So 0.13moles of Cu₂O will combine with 0.13 x 2 mole of NH₃
= 0.26moles of NH₃
Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;
Mass of N₂;
Mass = number of moles x molar mass
1 mole of Cu₂O will produce 1 mole of N₂
0.13 mole of Cu₂O will produce 0.13 mole of N₂
Mass = 0.13 x (2 x 14) = 3.64g
Answer:
The amount of work done on the system is 18234 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.
Explanation:
Thermodynamic work is called the transfer of energy between the system and the environment by methods that do not depend on the difference in temperatures between the two. When a system is compressed or expanded, a thermodynamic work is produced which is called pressure-volume work (p - v).
The pressure-volume work done by a system that compresses or expands at constant pressure is given by the expression:
W system= -p*∆V
Where:
- W system: Work exchanged by the system with the environment. Its unit of measure in the International System is the joule (J)
- p: Pressure. Its unit of measurement in the International System is the pascal (Pa)
- ∆V: Volume variation (∆V = Vf - Vi). Its unit of measurement in the International System is cubic meter (m³)
In this case:
- p= 10 atm= 1.013*10⁶ Pa (being 1 atm= 101325 Pa)
- ΔV= 2 L- 20 L= -18 L= -0.018 m³ (being 1 L=0.001 m³)
Replacing:
W system= -1.013*10⁶ Pa* (-0.018 m³)
Solving:
W system= 18234 J
<u><em>The amount of work done on the system is 18234 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.</em></u>
Answer:
The most abundant isotope is 1.007 amu.
Explanation:
Given data:
Average atomic mass = 1.008 amu
Mass of first isotope = 1.007 amu
Mass of 2nd isotope = 2.014 amu
Most abundant isotope = ?
Solution:
First of all we will set the fraction for both isotopes
X for the isotopes having mass 2.014 amu
1-x for isotopes having mass 1.007 amu
The average atomic mass is 1.008 amu
we will use the following equation,
2.014x + 1.007 (1-x) = 1.008
2.014x + 1.007 - 1.007 x = 1.008
2.014x - 1.007x = 1.008 - 1.007
1.007 x = 0.001
x= 0.001/ 1.007
x= 0.0009
0.0009 × 100 = 0.09 %
0.09 % is abundance of isotope having mass 2.014 amu because we solve the fraction x.
now we will calculate the abundance of second isotope.
(1-x)
1-0.0009 = 0.9991
0.9991 × 100= 99.91%
