Hess\'s Law Practice Given the following equations and of reaction (kJ) at 298 K

Question

Hess's Law Practice Given the following equations and of reaction (kJ) at 298 K for the reaction: 1. AH values, determine the heat 2 SO:(g) +2 P(s) +5 Chg)-2 soCI(l)+ 2 POCI() H./kJ . +10.3 7(a) +3/2 4 Hc1 (g) + Cla (g) ?PC1, (2) o,(g) ? 2 Cli (g) + 2 HO (1) Age/kJ -202.6 . 2. Calculate the heat of combustion (kJ) of propane, CsHs using the listed standard enthapy of reaction data CsHs(g)+508)3 Co:(g) +4 H20(g) c(a) O2 (g) co, (g) An/ks-393.5 3. Given the following equations and AHe values given below, determine the heat of reaction at 298 K for the reaction: 2 N2(8) +5 02(g)-2 N2Os(g) 240s (g) +H,0 (1) ? 2RNO, (1) AH/ -73.7 4. Given the following equations and reaction at 298 K for the reaction: values, determine the heat of C(s) +2 H2(g)-CHg) c(a) +02 (g)co2 (g) Hz(g) + 1/2O3(g) ?H,0 (1) co(g)+2 HO(1)CH(g)+2 0, (g) AHo/K-393.5 AHo/K-285.8 A/k+890.3

Explanation / Answer

1)

Lets number the reaction as 0, 1, 2, 3, 4 from top to bottom
required reaction should be written in terms of other reaction
This is Hess Law
required reaction can be written as:
reaction 0 = -2 * (reaction 1) +2 * (reaction 2) +2 * (reaction 3) -1 * (reaction 4)

So, ?Ho rxn for required reaction will be:
?Ho rxn = -2 * ?Ho rxn(reaction 1) +2 * ?Ho rxn(reaction 2) +2 * ?Ho rxn(reaction 3) -1 * ?Ho rxn(reaction 4)
= -2 * (10.3) +2 * (-325.7) +2 * (-306.7) -1 * (-202.6)
= -1082.8 KJ
Answer: -1082.8 KJ

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