Given Delta_fH deg/kJ mol^-1: C2H5OH(l) = -277.7, CO2(g) = -393.5, H2O(l) = -285.8. Select the correct statement(s).

The standard enthalpy of combustion of ethanol(l) to CO2(g) and H2O(l) is -1366.7 kJ mol^-1

The standard enthalpy of combustion of H2(g) equals Delta_fH deg(H2O(l))

The standard enthalpy of combustion of ethanol(l) to CO2(g) and H2O(l) is -1366.7 kJ mol^-1

The standard enthalpy of combustion of H2(g) equals Delta_fH deg(H2O(l))

Burning 4.6 g of ethanol (M = 46) releases 1366.7 kJ

The standard enthalpy of combustion of diamond equals Delta_fH deg(CO2(g))

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3td087

Given $Δ_{f} H^{\circ} / kJ mo l^{- 1}$ : $C X_{2} H X_{5} OH (l) = - 277.7$ , $CO X_{2} (g) = - 393.5$ , $H X_{2} O (l) = - 285.8$ . Select the correct statement(s).

AThe standard enthalpy of combustion of ethanol(l) to $CO X_{2} (g)$ and $H X_{2} O (l)$ is $- 1366.7 kJ mo l^{- 1}$
BThe standard enthalpy of combustion of $H X_{2} (g)$ equals $Δ_{f} H^{\circ} (H X_{2} O (l))$
CBurning $4.6 g$ of ethanol ( $M = 46$ ) releases $1366.7 kJ$
DThe standard enthalpy of combustion of diamond equals $Δ_{f} H^{\circ} (CO X_{2} (g))$

Δ_{c} H = 2 (- 393.5) + 3 (- 285.8) - (- 277.7) = - 1366.7 kJ mo l^{- 1}

, so the combustion value of ethanol quoted above is correct.

H X_{2} + \frac{1}{2} O X_{2} H X_{2} O (l)

is both the combustion of

H X_{2}

and the formation of liquid water. The formation reaction of

CO X_{2}

starts from graphite, not diamond. Burning

0.100 mol

ethanol releases only

136.7 kJ

, not

1366.7 kJ