Answer:
A. ΔG° = 132.5 kJ
B. ΔG° = 13.69 kJ
C. ΔG° = -58.59 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.
ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)
where,
n: moles
ΔH°f: standard enthalpy of formation
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))
ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)
ΔH° = 178.3 kJ
We can calculate the standard entropy of the reaction (ΔS°) using the following expression.
ΔS° = ∑np . S°p - ∑nr . S°r
where,
S: standard entropy
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))
ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)
ΔS° = 160.6 J/K. = 0.1606 kJ/K.
We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.
ΔG° = ΔH° - T.ΔS°
where,
T: absolute temperature
<h3>A. 285 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ
<h3>B. 1025 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ
<h3>C. 1475 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ
Answer: 2 Na (s) + Cl(g) -> 2 NaCl (s)
Explanation:
The empirical formula of the initial zinc oxide is ZnO.
<h3>What is Empirical Formula?</h3>
The empirical formula of a compound represents the ratios of elements in a compound but not the actual numbers or arrangement of the atoms.
It is the lowest whole number ratio of the element in the compound.
<h3>How to find out the empirical formula?</h3>
- Find out the given masses and molar masses of the elements
The molar mass of Zn = 65 gmol⁻¹
Given the mass of Zn = 2.156 g
The molar mass of Oxygen = 16 gmol⁻¹
The mass of Oxygen = Mass of a sample of zinc oxide - the mass of zinc metal
= (2.684 - 2.156) g
= 0.528 g
- Find the number of moles of the elements in the compound
The number of moles is given by
where m = given mass and
M = Molar mass
Number of moles of Zinc = 
= 0.033 moles
Number of moles of Oxygen =
= 0.033 moles
- Find the simplest ratios of the elements in the compound. To find the ratios simply divide the number of moles by the lowest number of moles obtained.
Here, the number of moles is the same for both elements. Hence, the simplest ratio for Zn:O is 1:1.
Therefore, the empirical formula of zinc oxide is ZnO.
Learn more about the empirical formula:
brainly.com/question/1603500
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Answer:
Gallium-72
Explanation:
The elements are identified by the number of protons of the atom, which is its atomic number.
In this case the number of protons 39 (atomic number 39) permit you to identify the element as gallium.
Now, to identify the isotope you tell the name of the element and add the mass number.
The mass number is the sum of the protons and the neutrons
In this case, the number of neutrons is the original 39 plus the 2 added suddenly, i.e. 39 + 2 = 41, so the mass number is 31 + 41 = 72
Therefore, the isotope is gallium - 72.
Answer:
Explanation:
For the reaction
C2H5OH (l) + 3 O2(g) = 2CO2(g) + 3 H2O
We can calculate the standard molar enthalpy of combustion using the standard enthalpies of formation of the species involved in the reaction according to Hess law:
ΔHºc = 2ΔHºf CO2 (g) + 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) - 3ΔHºfO2 (g) )
( we were not give the water state but we know we are at standard conditions so it is in its liquid state )
The ΔHºfs can be found in appropiate reference or texts.
ΔHºc = 2ΔHºf CO2 (g)+ 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) -+3ΔHºfO2 (g) )
= [ 2 ( -393.52 ) + 3 ( -285.83 ) ] - [( -276.2 + 0 ) ] kJ
ΔHºc = -1368.33 kJ
