Question

Part A

A+B→2C,ΔH1
1/2A+1/2B→C,ΔH2=?

Express your answer in terms of ΔH1.

ΔH2 =

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Part B

A+1/2B→C,ΔH1
2C→2A+B,ΔH2=?

Express your answer in terms of ΔH1.

ΔH2 =

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Part C

A→2B+2C,ΔH1
2A→4B+4C,ΔH2=?

Express your answer in terms of ΔH1.

ΔH2 =

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Part D

Calculate ΔHrxn for the reaction:

Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)

given these reactions and their ΔH′s:

2Fe(s)+32O2(g)→Fe2O3(s) ΔH=−824.2kJ

CO(g)+12O2(g)→CO2(g) ΔH=−282.7kJ

ΔHrxn =		kJ

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Part E

Calculate ΔHrxn for the reaction

CaO(s)+CO2(g)→CaCO3(s)

given these reactions and their ΔH values:

Ca(s)+CO2(g)2Ca(s)++12O2(g)O2(g)→→CaCO3(s),2CaO(s),ΔH=−814.1 kJΔH=−1269.8 kJ

Express the enthalpy in kilojoules to one decimal place.

ΔHrxn =		kJ

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Answer 1

2 2 # A+B C DH above equation divide with '2' + A+ + B - C AH AH v ® At 2 B - C AH, Reverse the above reaction and multiply with 2 sign of enthalpy change changes from (tve To ove 2c - ZA+B AH = selt -LAH, AH = -2 AH, A - 2B+22 AH, above equation multiply with 2 ZA~ 4B+40 AH = 2DH, Feo + 300 (g) -> 2Fe + 360 AH = ? eFe + 30 FeO3. DH = - 824.2kJ Reverse Ree the above reaction sign of AH changes from ove to the Fesz → 25%,+B sHa 824-2 kJ →0 co + 2 0 2 con AH = - 282.2 kJ above equation multiply with '3' 30+ 30 →300, AHO - 848.1 +5

0 + FRO 300 g + 2 2Fe + 3 y 300 t = 8.4.2 kJ t = -848. 1 kJ 300g) (9) (9) Feb + 3 co - 2Fe + 3 co. is 19) () (9) AH AH + AH =~23.9k5 u 290, +.com > cacos con AHF? (not clear) cals) * com + + 02 caro AH = - 814.1kJ 70 © 2cą + 2 - 2 cao, AH = -1269.8 kJ Reverse the above reaction and divide with 2 sign of Ato changes from sve to love +634.9 KJ - - 090 ( ) cal & AH = -814.1 kJ co + 2/19) (s) cal & co + 2 % ₂ cacos 15) 49) CaO > 4 + x 2 + 2 + (3 7 cao, t co - (aco, 192 + oH = -179.2 kJ