mistake archiever question

[jimmy1fernan]

---

Members don't see this ad.

Given ∆Hf for H2O(l) is –285.9 kJ/mol and ∆Hf for SO2(g) is –296.9 kJ/mol, calculate the enthalpy of formation, ∆Hf for H2S(g).

H2S(g) + 3/2O2(g) à H2O(l) + SO2(g) ∆H = - 562.6 kJ


<B> I thought it was A not B!! It can't be B. is it me or there is a mistake with this question? please help__ </B>


A.+ 20.2 kJ/molB.– 20.2 kJ/molC.+ 1145.5 kJ/molD.– 1145.5 kJ/molE.0 kJ/mol



&#8710;H = &#931; &#8710;Hf(products) - &#931;<SPAN style="FONT-FAMILY: Arial"> &#8710;Hf(reactants)

 

[jimmy1fernan]

A. =20.2
B.=-20.2
C.=+1145.5
D.=-1145.5

these are the answer choices

A.+ 20.2 kJ/molB.– 20.2 kJ/molC.+ 1145.5 kJ/molD.– 1145.5 kJ/molE.0 kJ/mol

 

[KTDMD]

Given &#8710;Hf for H2O(l) is –285.9 kJ/mol and &#8710;Hf for SO2(g) is –296.9 kJ/mol, calculate the enthalpy of formation, &#8710;Hf for H2S(g).

H2S(g) + 3/2O2(g) à H2O(l) + SO2(g) &#8710;H = - 562.6 kJ


<B> I thought it was A not B!! It can't be B. is it me or there is a mistake with this question? please help__ </B>


A.+ 20.2 kJ/molB.– 20.2 kJ/molC.+ 1145.5 kJ/molD.– 1145.5 kJ/molE.0 kJ/mol



&#8710;H = &#931; &#8710;Hf(products) - &#931;<SPAN style="FONT-FAMILY: Arial"> &#8710;Hf(reactants)

You forgot to move the negative sign:

- &#931;&#8710;Hf(reactants) = 20.2 kJ/mol
&#931;&#8710;Hf(reactants) = -20.2 kJ/mol

Hope this helps.