Without doing any calculations, determine the sign < or> for each system and sur

Question

Without doing any calculations, determine the sign < or> for each system and surrounding for each chemical reactions.

2NO2(g)--->2NO(g)+O2(g) change in Hrxn +113.1 kJ

A) Change S sys. > 0, change S surr. > 0

B) Change S sys. < 0, change S surr. > 0

c) Change S sys. > 0, change S surr. < 0

D) Change S sys. < 0, change S surr. < 0

Please explain in detail why for high rating, thanks

Explanation / Answer

(A) N2(g) + O2(g) ------> 2NO(g) ?H = +180.7kJ (B) 2NO(g) + O2(g) ------> 2NO2(g) ?H = -113.1kJ (C) 2N2O(g) ------> 2N2(g) + O2(g) ?H = -163.2kJ only by using 1/2 of equation (C) can we get 1 N2O as a reactant only by using 1/2 of the reverse of (B) can we get 1 NO2 as a reactant 1/2(C): 1N2O(g) ------> 1N2(g) + 1/2 O2(g -1/2(B): 1NO2(g) ---> 1NO(g) + 1/2 O2(g) this gives us our two starting materials: N2O(g) + NO2(g) ------> 1NO(g) & 1N2 & 1 O2 only by using 1(A) will we throw in 1N2 & 1 O2 as reactants to cancel out the 1N2 & 1 O2 already there as products: 1/2(C) -1/2(B) gave : N2O(g) + NO2(g) ------> 1NO(g) & 1N2 & 1 O2 1(A) gives : N2(g) + O2(g) ------> 2NO(g not only did that step cancel out the N2(g) + O2(g) , but it gave us the extra 2NO(g that we needed as products, leaving us with: N2O(g) + NO2(g) ------> 3NO(g) Hess's law says that if 1/2(C) -1/2(B) &1(A) gave us the reaction that we want,.... then dH of 1/2(C) - dH of 1/2(B) & dH of 1(A) will give us the dH we want dHrxn = dH of 1/2(C) - dH of 1/2(B) & dH of 1(A) dH rxn = (-81.6) & -(-56.55) & (+180.7) dH rxn = -81.6 + 56.55 + 180.7 dHrxn = +155.65 kJ dHrxn = 155.7 kJ

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