For the reaction below, the constant pressure heat of reaction is qp = −2451 kJ mol−1 at 25 °C. What is the constant volume heat of reaction, qV, at 25 °C?

12 CO(g) + 25 H2(g) ⟶ C12H26(l) + 12 H2O(l)

Enter your answer in kJ mol−1

dH - (delta n)RT = ?

To find the constant volume heat of reaction, qV, we can use the equation:

qV = qp + ΔnRT

Where:
- qp is the constant pressure heat of reaction
- Δn is the change in the number of moles of gas
- R is the ideal gas constant (8.314 J mol−1 K−1)
- T is the temperature in Kelvin

First, let's calculate Δn, the change in the number of moles of gas:

Δn = (moles of products) - (moles of reactants)

From the balanced equation, we can see that the moles of products are 1 (C12H26(l) + 12 H2O(l)).
The moles of reactants are 12 (CO(g)) + 25 (H2(g)) = 37.

Therefore, Δn = 1 - 37 = -36.

Now we can substitute the values into the equation:

qV = -2451 kJ mol−1 + (-36 mol) * (8.314 J mol−1 K−1) * (25 °C + 273.15 K)

Note:
- We converted the temperature from Celsius to Kelvin by adding 273.15.

Calculating:

qV = -2451 kJ mol−1 + (-36 mol) * (8.314 J mol−1 K−1) * (298.15 K)

qV = -2451 kJ mol−1 - 8878.352 kJ mol−1

qV = -11329.352 kJ mol−1

Therefore, the constant volume heat of reaction, qV, at 25 °C is -11329.352 kJ mol−1.

To find the constant volume heat of reaction, qV, at 25 °C, we need to determine the relationship between heat at constant pressure (qp) and heat at constant volume (qV).

The relationship between qV and qp can be expressed using the ideal gas law equation:
qV = qp - nRΔT

Where:
qV is the constant volume heat of reaction,
qp is the constant pressure heat of reaction,
n is the total number of moles of gas involved in the reaction,
R is the gas constant (8.314 J mol−1 K−1), and
ΔT is the temperature change in Kelvin.

First, we need to convert the given temperature from Celsius to Kelvin.
25 °C + 273.15 = 298.15 K

Next, we need to determine the number of moles of gas involved in the reaction. From the balanced equation, we can see that there are 12 moles of CO and 25 moles of H2 involved.

Now, we can calculate the total number of moles of gas:
n = 12 moles of CO + 25 moles of H2
n = 37 moles

Substituting the values into the equation, we get:
qV = -2451 kJ mol−1 - (37 mol)(8.314 J mol−1 K−1)(298.15 K)

Now we can calculate qV:
qV = -2451 kJ mol−1 - (37 mol)(8.314 J mol−1 K−1)(298.15 K)
qV = -2451 kJ mol−1 - (37)(8.314)(298.15) J K−1
qV = -2451 kJ mol−1 - 93009.449 J
qV = -93246009.449 J

To convert J to kJ, we divide by 1000:
qV = -93246009.449 J / 1000
qV = -93246.009 kJ

Therefore, the constant volume heat of reaction, qV, at 25 °C is approximately -93246.009 kJ mol−1.