Question

6. Given the following thermochemical reactions 2NO N,+O 2NO+O,2NO AH=-180 kJ AH=-112 kJ Calculate the enthalpy change of the reaction of nitrogen with oxygen to produce nitrogen dioxide. N,+20,2NO, AH r? Lo

Answer 1

9.6: Given - 2 NO - > N2 + O2 AH = - 180 KJ - 2 N0 +02 - 2 NO2 4H = - 112 KJ @ Now, Writing equations ② in opposition direction, we will get N2 + 0, >2NO AH= + 180 KJ - 3 Now, doing equation 3 + equation ©, we will get- N2 + O2 -> 2N0 AH = + 180 kJ 2NO + 02 - 2 NO2 AH = -112 KI N2 + 202 --> 2NO, 4Hrxu. = (4H + 4H) : A Horreu. = (4H + 4H) = +180 KJ + (-112 KJ) = + 180 kJ – 112 KJ = 68 KJ, Ahren value will be + 68 KJ. Am.

YA Q.7: We know that - A Hoxu. = E n4H50 ( Products) - En Hj (reactants) Given the neaction - C₂ HG OH (1) + O2(q) - CH₂ CO₂H(A) + H₂O (1) :: Ahren. = {1x 44;"(CM3c0,4}+{ıxan;(.)}]- [{ixam (C, 15 ow)}+{ıxan; (o? Given- Ahi of C₂H5OH (1) = - 277.7 kJ/mol AHf of CH₂ CO₂H (2) = - 484.5 kJ/mol AHY of H20 (4) = -285.8 kJ/mol Hij of O2(g) = 0 KJ/mos. O LEO

:: Alman = {vanj (en czw)tfix.ang (4]– Bixang (.313 0w33 +{sxanj (o,}] ={{ix (-434-5)}+{1x =285-8}]+[{ix (-277-7)}+{ox 0}] = [- 484.5 + (-285-8)]-[[277.7) +o] = (-484-5 - 285.8] + 277.7 = -770.3 +277.7 = - 492.6 kJ/mol .. A Hoxn value will be - 492.6 kJ/mol. Aug.