At 25 °C , for the combustion of 1 mole of liquid benzene, the heat of reaction at constant pressure is given by C6H6 (l) +(7/2) O2(g) → 6CO2(g) + 3H2O(l) , Δ H= 780980 cal Calculate the heat of reaction at constant volume

Question

At 25 °C , for the combustion of 1 mole of liquid benzene, the heat of reaction at constant pressure is given by C6H6 (l) +(7/2) O2(g) → 6CO2(g) + 3H2O(l) , Δ H= 780980 cal Calculate the heat of reaction at constant volume (A)  780.086 kcals  (B)  -780.086 kcals  (C)  -390.043 kcals  (D) 779.49 kcal

Tardigrade Admin · Accepted Answer

Δ H =Δ U+Δ ng RT (where, Δ H= heat of reaction at constant pressure Δ U= heat of reaction at constant volume) heat of reaction at constant Δ ng=6-(7/2)=(5/2)=2.5 ∴ Δ U=Δ H - Δ ng RT =780980-2.5×2×298 =780980-1490=779490 =779.49 kcal

Q. At $2 5^{\circ} C$ , for the combustion of $1$ mole of liquid benzene, the heat of reaction at constant pressure is given by
$C_{6} H_{6} (l) + \frac{7}{2} O_{2} (g) \to 6 C O_{2} (g) + 3 H_{2} O (l)$ , $Δ H = 780980 c a l$
Calculate the heat of reaction at constant volume

$780.086 k c a l s$

$- 780.086 k c a l s$

$- 390.043 k c a l s$

$779.49 k c a l$

Q. At 25∘C , for the combustion of 1 mole of liquid benzene, the heat of reaction at constant pressure is given by C6​H6​(l)+27​O2​(g)→6CO2​(g)+3H2​O(l) , ΔH=780980cal Calculate the heat of reaction at constant volume

780.086kcals

−780.086kcals

−390.043kcals

779.49kcal

ΔH=ΔU+Δng​RT (where, ΔH= heat of reaction at constant pressure ΔU= heat of reaction at constant volume) heat of reaction at constant Δng​=6−27​=25​=2.5 ∴ΔU=ΔH−Δng​RT =780980−2.5×2×298 =780980−1490=779490 =779.49kcal

Q. At $2 5^{\circ} C$ , for the combustion of $1$ mole of liquid benzene, the heat of reaction at constant pressure is given by
$C_{6} H_{6} (l) + \frac{7}{2} O_{2} (g) \to 6 C O_{2} (g) + 3 H_{2} O (l)$ , $Δ H = 780980 c a l$
Calculate the heat of reaction at constant volume

$780.086 k c a l s$

$- 780.086 k c a l s$

$- 390.043 k c a l s$

$779.49 k c a l$