Billions of years ago, Earth was a very hot place due to

Given the following values for the change in enthalpy (deltaH) and entropy (deltaS), which of the following processes can occur

Shtirlitz [24]

Answer:

Option A and B

Explanation:

(a) DeltaH = -84 kj mol-2 (-20 kcal mol-1), DeltaS = +125j mol-2K-1)(+30 cal mol-1 K-1)

Delta G = Delta H – T * DS

Substituting the given values, we get –

Delta G = -84 -298 *(125/1000) = -121.25 KJ/mol

Delta G is negative hence the process is spontaneous and will not violate the second law of thermodynamics

(b) DeltaH = -84 kj mol-2 (-20 kcal mol-1), DeltaS = -125j mol-2K-1)(-30 cal mol-1 K-1)

Delta G =-84 -298 *(-125/1000) = -46.75 KJ/mol

Delta G is negative hence the process is spontaneous and will not violate the second law of thermodynamics

(c) DeltaH = +84 kj mol-2 (+20 kcal mol-1), DeltaS = +125j mol-2K-1)(+30 cal mol-1 K-1)

Delta G = 84 -298 *(125/1000) = +46.75 KJ/mol

Delta G is positive hence the process is non-spontaneous and will violate the second law of thermodynamics

(d) DeltaH = +84 kj mol-2 (+20 kcal mol-1), DeltaS = +125j mol-2K-1)(-30 cal mol-1 K-1)

Delta G = 84 -298 *(-125/1000) = + 121.25 KJ/mol

Delta G is positive hence the process is non-spontaneous and will violate the second law of thermodynamics

6 0

3 years ago

Use the following information S(s)+O2(g)→SO2(g), ΔH∘ = -296.8kJ SO2(g)+12O2(g)→SO3(g) , ΔH∘ = -98.9kJ to calculate ΔH∘f for H2SO

Irina-Kira [14]

Answer : The enthalpy of formation of $H_2SO_4$ is, -812.4 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation of $H_2SO_4$ will be,

$S+H_2+2O_2\rightarrow H_2SO_4$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $S+O_2\rightarrow SO_2$ $\Delta H_1=-298.2$

(2) $SO_2+\frac{1}{2}O_2\rightarrow SO_3$ $\Delta H_2=-98.2$

(3) $SO_3+H_2O\rightarrow H_2SO_4$ $\Delta H_3=-130.2$

(4) $H_2+\frac{1}{2}O_2\rightarrow H_2O$ $\Delta H_4=-285.8$

Now adding all the equations, we get the expression for enthalpy of formation of $H_2SO_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4$

$\Delta H=(-298.2)+(-98.2)+(-130.2)+(-285.8)$

$\Delta H=-812.4kJ/mol$

Therefore, the enthalpy of formation of $H_2SO_4$ is, -812.4 kJ/mole

3 0

3 years ago

10) Calculate the energy change for the formation of LiCl(s) from its elements in their standard states and the following tabula

marissa [1.9K]

Answer:

-400.3 kJ/mol

Explanation:

3 0

3 years ago

According to Hund's Rule, if you were filling the three orbitals at the 2p energy level, you would first put an electron in the

GenaCL600 [577]

Answer:

True.

Explanation:

The Hund's Rule states that all orbitals must be singled occupied before any orbital is doubly occupied, and all the electrons at the singly occupied orbitals have the same spin number. By doing that, the electrons filled the lowest energy orbitals first.

The 2p level has 3 orbitals: 2px, 2py, and 2pz. So, when filling it, first put an electron in the 2px, then in the 2py, then and the 2pz (all with the same spin). After that, the remains electrons can be paired up.

5 0

3 years ago

A substance has a high melting point and conducts electricity in the liquid phase The is substance is

posledela

Answer:

oxygen

Explanation:

?

3 0

3 years ago