Chemistry help! Please show your work and explaination on how you calculated the

Question

Chemistry help! Please show your work and explaination on how you calculated the answer! Please and ThankYou Questions 33-35 Consider the reaction: 4NH3(g) 702 (g) 4NO2(g) 6H20 4H NO2 (g) 34 KJ/mol 4H INH3(g) --46 KJ/mol 33. AH at 25 C is a. 1396 KJ b. 2688 KU/mo c.. 1396 KU d. 206 KJ e, 3172 KJ 34 AE is a. 1379 KU b.. 1379 KU c. 2705 KJ d. 2705 KU e. 14655 KU 35. Is this reaction a. Endothermic b. Exothermic c. Both 36. Consider i. 2B (s) 3/2 O2 (g) B203(s) AH--1273 KU ii. B2H6(g) 302 (g) B203(s) +3H20(g) AH -2035 KJ AH -286 KU 4H 44 KU iv. H20 H20(g) AH for 2B(s) 3H2(g) B2H6(g) must be a. 36 KU b. -36 KU

Explanation / Answer

33. Answer:-a

We know that

Enthalpy change of a reaction=H=sum of enthalpy of formation of products-sum of enthalpy of formation of reactants

So H=4(34)+6(-286)-4(-46)=-1396KJ/mol

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