Question

The heat of combustion of solid benzoic acid at constant volume is 321.30 k j at 27 c 321.30kjat27c . The heat of combustion at constant pressure is

Answer 1

-321.30 - 150 R

C6H5COOH(s)+152O2(g)⟶7CO2(g)+3H2O(l)C6H5COOH(s)+152O2(g)⟶7CO2(g)+3H2O(l)

Since, Δng=nP−nRΔng=nP−nR

ΔngΔng = 7 - 152=−12152=−12

ΔH=ΔE+ΔngRTΔH=ΔE+ΔngRT

ΔH=−321.30−(12×R×300)=−321.30−150R

Answer 2

C6H5COOH(s) + 15/2 O2(g) ----> 7CO2(g) + 3H2O(l)

1 mol          15/2 mol    7 mol

△ng = np - nR = 7 - (15/2) = - 0.5 mol ;

qv = -312.3 KJ

qp = qv +  △ng RT ---(1)

qp = -312.3 + (-0.5) x R x

(273 + 27 i.e, 300K)

= -312.3 - 150 R

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