Answer : The correct option is, (1) 0
Explanation :
Oxidation state or oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.
Oxidation numbers are generally written with sign (+ and -) first and then the magnitude.
When the atoms are present in their elemental state then the oxidation number will be zero.
As per question, 'Mn' atom is a free element that means it is present in their elemental state then the oxidation state will be zero.
Hence, the correct option is, (1) 0
Answer:
It is greater than -600kJ/mol and the amount of energy required to break bonds is greater than the amount of energy released in forming bonds.
Explanation:
In an endothermic reaction, the reaction requires a determined amount of energy to occurs.
The reaction of the problem has H = -600kJ/mol. The reaction is endothermic and the energy that the reaction needs is absorbed by the reactants. That means, the energy of products:
Is greater than -600kJ/mol and the amount of energy required to break bonds is greater than the amount of energy released in forming bonds.
<u>Answer:</u> The equilibrium constant for this reaction is 
<u>Explanation:</u>
The equation used to calculate standard Gibbs free change is of a reaction is:
![\Delta G^o_{rxn}=\sum [n\times \Delta G^o_{(product)}]-\sum [n\times \Delta G^o_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_%7B%28reactant%29%7D%5D)
For the given chemical reaction:
The equation for the standard Gibbs free change of the above reaction is:
![\Delta G^o_{rxn}=[(1\times \Delta G^o_{(Ni(CO)_4(g))})]-[(1\times \Delta G^o_{(Ni(s))})+(4\times \Delta G^o_{(CO(g))})]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20G%5Eo_%7B%28Ni%28CO%29_4%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20G%5Eo_%7B%28Ni%28s%29%29%7D%29%2B%284%5Ctimes%20%5CDelta%20G%5Eo_%7B%28CO%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta G^o_{rxn}=[(1\times (-587.4))]-[(1\times (0))+(4\times (-137.3))]\\\\\Delta G^o_{rxn}=-38.2kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-587.4%29%29%5D-%5B%281%5Ctimes%20%280%29%29%2B%284%5Ctimes%20%28-137.3%29%29%5D%5C%5C%5C%5C%5CDelta%20G%5Eo_%7Brxn%7D%3D-38.2kJ%2Fmol)
To calculate the equilibrium constant (at 58°C) for given value of Gibbs free energy, we use the relation:
where,
= Standard Gibbs free energy = -38.2 kJ/mol = -38200 J/mol (Conversion factor: 1 kJ = 1000 J )
R = Gas constant = 8.314 J/K mol
T = temperature = ![58^oC=[273+58]K=331K](https://tex.z-dn.net/?f=58%5EoC%3D%5B273%2B58%5DK%3D331K)
= equilibrium constant at 58°C = ?
Putting values in above equation, we get:
Hence, the equilibrium constant for this reaction is
The crude oil is heated in a process of fractional distillation (note: you can research this online if you require more detail) and this process causes different hydrocarbons to condense at different heights up the distillation column. They will therefore run down the side of the column and into different areas depending on their boiling point.
