Nitrogen (N2) and hydrogen (H2) gases react to form ammonia, which requires -99.4 J/K of standard entropy (ΔS°).
What is standard entropy?
The difference between the total standard entropies of the reaction mixture and the summation of the standard entropies of the outputs is the standard entropy change. Each entropy in the balanced equation needs to be compounded by its coefficient, as shown by the letter "n."
Calculation:
Balancing the given reaction following-
1/2 N₂(g) + 3/2 H₂ (g)→ NH₃ (g)
ΔS° = [1 mol x S° (NH₃)g] - [1/2 mol x S° (N₂)g] - [3/2 mol x S°(H₂)g]
Here S° = standard entropy of the system
Insert into the aforementioned equation all the typical entropy values found in the literature:
ΔS° = [1 mol x 192.45 J/mol.K] - [1/2 mol x 191.61 J/mol.K] - [3/2 mol x 130.684 J/mol.K]
⇒ΔS° = - 99.4 J/K
Therefore, the standard entropy, ΔS° is -99.4 J/K.
Learn more about standard entropy here:
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1.806x10^24
Written equation form(always start the equation off with what you know based off of the question!):
3mol(CCl4)•6.022x10^23/1mol = 1.806x10^24
Good luck!
I believe the correct answer is the first option. To increase the molar concentration of the product N2O4, you should increase the pressure of the system. You cannot determine the effect of changing the temperature since we cannot tell whether it is an endothermic or an exothermic reaction. Also, decreasing the number of NO2 would not increase the product rather it would shift the equilibrium to the left forming more reactants. The only parameter we can change would be the pressure. And, since NO2 takes up more space than the product increasing the pressure would allow the reactant to collide more forming the product.
Answer:
carbon dioxide concentration goes down, temperature goes down. Carbon dioxide goes up temp goes up, carbon dioxide is directly related to temperature by insulating it in the earths atmosphere and if there's less of it then the temp would go down.
Explanation: