Constant pressure heat of reaction -> constant volume q

[jybe]

Homework Statement

For the reaction below, the constant pressure heat of reaction is qp = −3256 kJ mol−1 at 25 °C. What is the constant volume heat of reaction, qV , at 25 °C?

16 CO(g) + 33 H2(g) ⟶ C16H34(l) + 16 H2O(l)

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

Homework Equations

At constant pressure, the heat of reaction is equal to the enthalpy change

At constant volume, the heat of reaction is equal to the internal energy change: ΔH = ΔU + ΔnRT

ΔU = ΔH - ΔnRT

The Attempt at a Solution

ΔU = -3256kJ/mol - ΔnRT
Edit:

ΔU = -3256kJ/mol - (-49)(8.314x10^-3)(298.15)

ΔU = -3135 kJ/mol

Have I done it correctly?

 

[Chestermiller]

It looks correct to me.

 

[BvU]

It looks correct to me.

I was wondering about the -49 used for $\Delta n$ -- what is the reasoning ?

 

[jybe]

I was wondering about the -49 used for $\Delta n$ -- what is the reasoning ?

It's the change in moles of gas gathered from the equation, used to calculate work done