Question

According to the equation, C6H6(l)+7/2O2(g)−→−6CO2(g)+3H2O(l); ΔH=−x kJ The energy evolved when 3.9 gm of benzene is burnt in air is -163.2 KJ, heat of combustion of benzene is (A) 32.46 KJ (B) 16.32 KJ (C) 326.4 KJ (D) −3264 KJ

Answer 1

Answer:

$c6h6 \: + \frac{7}{2} h2o \: = 6co2 \: + 3h2o$

δH = -x KJ

Molecular weight of benzene = 78g

The energy evolved when 3.9 gm of benzene is burnt in air is = ( -163.2 KJ )

Energy evolved in 1 g combustion = ( -163.2 / 3.9 ) KJ

Hence energy evolved in 78 gm of benzene

$\frac{ - 163.2}{3.4} \times 78 \\ \\ - 3264 \: kj$

Hence heat of combustion of benzene is - 3264 KJ

Thanks

Answer 2

Answer:

Explanation:C6H6 __78gms

78gms=1 mole

3.9gms of C6H6= -163.2kJ

78gms ofC6H=

x=32.64kJ