Question

the heat up combustion of benzene is -3268× 10^3 Joule at 27 degrees Celsius at constant pressure what is it's value at constant volume and at same temperature C6H6(l)+15/2 O2(g)----->6CO2(g)+ 3H2O(l)​

Answer 1

Answer:

∆H = ∆E + ∆ng RT

∆E = ∆H - ∆ng RT

∆H = -3268 × 10^3 Joule

∆ng = 6-7.5 = -1.5

R = 8.3

T = 300K

∆E = ∆H + 1.5 × 8.3 × 300

Solve get answer

My Aim is Learn about Concept

This is mathematics