<u>Answer:</u> The value of
of the reaction is 28.38 kJ/mol
<u>Explanation:</u>
For the given chemical reaction:

- The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28reactant%29%7D%5D)
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SO_2Cl_2(g))})]-[(1\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(Cl_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_2Cl_2%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_2%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28Cl_2%28g%29%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-364))]-[(1\times (-296.8))+(1\times 0)]=-67.2kJ/mol=-67200J/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-364%29%29%5D-%5B%281%5Ctimes%20%28-296.8%29%29%2B%281%5Ctimes%200%29%5D%3D-67.2kJ%2Fmol%3D-67200J%2Fmol)
- The equation used to calculate entropy change is of a reaction is:
![\Delta S^o_{rxn}=\sum [n\times \Delta S^o_f_{(product)}]-\sum [n\times \Delta S^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20S%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20S%5Eo_f_%7B%28reactant%29%7D%5D)
The equation for the entropy change of the above reaction is:
![\Delta S^o_{rxn}=[(1\times \Delta S^o_{(SO_2Cl_2(g))})]-[(1\times \Delta S^o_{(SO_2(g))})+(1\times \Delta S^o_{(Cl_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20S%5Eo_%7B%28SO_2Cl_2%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20S%5Eo_%7B%28SO_2%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20S%5Eo_%7B%28Cl_2%28g%29%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta S^o_{rxn}=[(1\times 311.9)]-[(1\times 248.2)+(1\times 223.0)]=-159.3J/Kmol](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20311.9%29%5D-%5B%281%5Ctimes%20248.2%29%2B%281%5Ctimes%20223.0%29%5D%3D-159.3J%2FKmol)
To calculate the standard Gibbs's free energy of the reaction, we use the equation:

where,
= standard enthalpy change of the reaction =-67200 J/mol
= standard entropy change of the reaction =-159.3 J/Kmol
Temperature of the reaction = 600 K
Putting values in above equation, we get:

Hence, the value of
of the reaction is 28.38 kJ/mol
Answer: 22 kJ amount of energy is released in the following reaction.
Explanation: There are two types of reaction on the basis of amount of heat absorbed or released.
1. Endothermic reactions: These are the type of reactions in which reactants absorb heat to form the products. The energy of the reactants is less than the energy of the products.
2. Exothermic reactions: These are the type of reactions in which heat is released from the chemical reactions. The energy of the products is less than the reactants.
Sign convention for
: This value is negative for exothermic reactions and positive for endothermic reactions.
For the given chemical reaction,
Energy of the products is less than the energy of the reactants, Hence, this reaction will be a type of exothermic reaction and energy will be released during this chemical change.
Amount of energy released = (350 - 372) kJ = -22kJ
Negative sign symbolizes the energy is being released. So, 22 kJ amount of energy is released in the following reaction.
Hey there!:
Given the reaction:
2 C2H2 + 5 O2 → 4 CO2 + 2 H2O
5 moles O2 ------------- 4 moles CO2
3.00 moles O2 ---------- ( moles of CO2 ?? )
moles of CO2 = 3.00 * 4 / 5
moles of CO2 = 12 / 5
moles of CO2 = 2.4 moles
So, molar mass CO2 = 44.01 g/mol
Therefore:
1 mole CO2 -------------- 44.01 g
2.4 moles CO2 ---------- ( mass of CO2 )
mass of CO2 = 2.4 * 44.01 / 1
mass of CO2 = 106 g
Answer A
Hope that helps!
Answer: Transition from X to Y will have greater energy difference.
Explanation: For studying the energy difference, we require Planck's equation.

where, h = Planck's Constant
c = Speed of light
E = Energy
= Wavelength of particle
From the equation, it is visible that the energy and wavelength follow inverse relation which means that with low wavelength value, energy will be the highest and vice-versa.
As electron A falls from X-energy level to Y-energy level, it releases blue light which has low wavelength value (around 470 nm) which means that it has high energy.
Similarly, Electron B releases red light when it falls from Y-energy level to Z-energy level, which has high wavelength value (around 700 nm), giving it a low energy value.
Energy Difference between X-energy level and Y-energy level will be more.