For the reaction below, the constant pressure heat of reaction is qp = −1249 kJ mol−1 at 25 °C. What is the constant volume heat of reaction, qV , at 25 °C?

6 CO(g) + 13 H2(g) ⟶ C6H14(l) + 6 H2O(l)

Enter your answer in kJ mol−1, rounded to the nearest kilojoule.

The constant volume heat of reaction is qV =

Question

For the reaction below, the constant pressure heat of reaction is qp = −1249 kJ mol−1 at 25 °C. What is the constant volume heat of reaction, qV , at 25 °C?

6 CO(g) + 13 H2(g) ⟶ C6H14(l) + 6 H2O(l)

Enter your answer in kJ mol−1, rounded to the nearest kilojoule.

The constant volume heat of reaction is qV =

rEsTiNpEaCe · Answer

@Vocaloid

rEsTiNpEaCe · Answer

I tried this: 
qV = -1249-(-19x8.314x298.15) = 45849 kj mol-1 
But it's saying my answer is wrong. Can you help please? I've been struggling a lot on this

rEsTiNpEaCe · Answer

P.S. I got my -19 from the amount of gas moles. Products - Reactants: 0-(6+13) = -19

rEsTiNpEaCe · Answer

@dude

Tranquility · Answer

I've actually never really seen this formula being used often, but you plugged in everything correctly. Where you went wrong is with the units.

R = 8.314 J/mol*K
The final answer has to be in kJ/mol so we can re-write 8.314 J/mol*K as 8.314*10^-3 kJ/mol*K

So simplify this
qV = -1249kJ/mol - (-19 * (8.314*10^-3 kJ/mol*K) * 298.15 K)