For which reaction, carried out at standard conditions, wouldboth the enthalpy and entropy changes drive the reaction in thesame direction? Please EXPLAIN. A. 2H2(g) + O2(g) --->2H2O(l) ΔH = -571.1 kJ

B. 2Na(s) + Cl2(g)>2NaCl(s) ΔH = -822.0kJ

C. N2(g) + 2O2(g) >2NO2(g) ΔH= +67.7 kJ

D. 2NH3(g) > N2(g) +3H2(g) ΔH= +92.4 kJ

2NH3(g) > N2(g) +3H2(g) ΔH= +92.4 kJ

Explanation:

Entropy increases with increase in the number of particles from left to right in a reaction. Hence, the reaction; 2NH3(g) > N2(g) +3H2(g) ΔH= +92.4 kJ is favoured by increase in entropy.

Similarly, the enthalpy change for the reaction is only +92.4 KJ. Hence for this reaction, both enthalpy and entropy changes drive the reaction in the same direction.

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons