If the molar heat of combustion of liquid benzene at constant volume and 300k is -3272KJ. Calculate the heat of combustion at constant pressure at thesame temperature

The heat at constant pressure is -3,275.7413 kJ

Explanation:

The combustion equation is 2C₆H₆ (l) + 15O₂ (g)  → 12CO₂ (g) + 6H₂O (l)

= (12 - 15)/2 = -3/2

We have;

Where R and T are constant, and ΔU is given we can write the relationship as follows;

Where;

H = The heat at constant pressure

U = The heat at constant volume = -3,272 kJ

= The change in the number of gas molecules per mole

R = The universal gas constant = 8.314 J/(mol·K)

T = The temperature = 300 K

Therefore, we get;

H = -3,272 kJ + (-3/2) mol ×8.314 J/(mol·K) ×300 K) × 1 kJ/(1000 J) = -3,275.7413 kJ

The heat at constant pressure, H = -3,275.7413 kJ.

it is a good conductor

explanation:

c. 75.7

i think that is the right