What will be the heat change at constant volume for the reaction whose heat change at constant pressure is -560 kcal at 27 °C ? The reaction is C8H16+12O2 →8CO2+8H2O (Given R = 2 cal mol-1 K-1 )

Question

What will be the heat change at constant volume for the reaction whose heat change at constant pressure is -560 kcal at 27 °C ? The reaction is C8H16+12O2 →8CO2+8H2O (Given R = 2 cal mol-1 K-1 ) (A)  -558200 calories (B)  442800 calories (C)  -561800 calories (D)  368240 calories

Tardigrade Admin · Accepted Answer

Δ H =-560 kcal, Δ E=? C8H16(l)+12O2(g) arrow 8CO2(g)+8H2O(l) Δ H=Δ E+Δ ng RT Δ E=Δ H-Δ ng RT =-560,000-(-4×2×300) =-560,000+2400 =-5,57,600 cal

Q. What will be the heat change at constant volume for the reaction whose heat change at constant pressure is $- 560 k c a l$ at $27^{\circ} C$ ? The reaction is
$C_{8} H_{16} + 12 O_{2} \to 8 C O_{2} + 8 H_{2} O$
(Given $R = 2 c a l m o l^{- 1} K^{- 1}$ )

$- 558200$ calories

$442800$ calories

$- 561800$ calories

$368240$ calories

$Δ H = - 560 k c a l$ , $Δ E = ?$
$C_{8} H_{16 (l)} + 12 O_{2 (g)} \to 8 C O_{2 (g)} + 8 H_{2} O_{(l)}$
$Δ H = Δ E + Δ n_{g} RT$
$Δ E = Δ H - Δ n_{g} RT$
$= - 560, 000 - (- 4 \times 2 \times 300)$
$= - 560, 000 + 2400$
$= - 5, 57, 600 c a l$

Q. What will be the heat change at constant volume for the reaction whose heat change at constant pressure is −560kcal at 27∘C ? The reaction is C8​H16​+12O2​→8CO2​+8H2​O (Given R=2calmol−1K−1 )

−558200 calories

442800 calories

−561800 calories

368240 calories