Answer:
4.7 kJ/kmol-K
Explanation:
Using the Debye model the specific heat capacity in kJ/kmol-K
c = 12π⁴Nk(T/θ)³/5
where N = avogadro's number = 6.02 × 10²³ mol⁻¹, k = 1.38 × 10⁻²³ JK⁻¹, T = room temperature = 298 K and θ = Debye temperature = 2219 K
Substituting these values into c we have
c = 12π⁴Nk(T/θ)³/5
= 12π⁴(6.02 × 10²³ mol⁻¹)(1.38 × 10⁻²³ JK⁻¹)(298 K/2219 K)³/5
= 9710.83(298 K/2219 K)³/5
= 1942.17(0.1343)³
= 4.704 J/mol-K
= 4.704 × 10⁻³ kJ/10⁻³ kmol-K
= 4.704 kJ/kmol-K
≅ 4.7 kJ/kmol-K
So, the specific heat of diamond in kJ/kmol-K is 4.7 kJ/kmol-K
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Exothermic releases energy(heat) from the system to the surrounding
endothermic takes energy (heat) from the surrounding to the system
The sign of the entropy change, ΔS, for the following processes include:
- I2 (s) + ½ Cl2 (g) ← → ICl (g) - positive
- 3 Ag (s) + 4 HNO3 (aq) → 3AgNO3 (aq) + NO (g) + 2 H2O (l) - positive
- Cl2 (g) → Cl2 (l) - negative
- C5H12 (g) + 8 O2 (g) → 5 CO2 (g) + 6 H2O (l) - negative
What is Entropy?
This is referred to the degree of randomness or disorderliness of a system and is denoted as S.
Entropy change is usually positive when solid or liquid reactant is converted into gas product and vice versa.
Read more about Entropy here brainly.com/question/419265
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Answer:
3
2
Explanation:
3O2 => 2O3
the equation is balancing
