Data in kJ mol^-1: Delta_cH(graphite) = -393.5; Delta_cH(CO) = -283.0; Delta_cH(H2) = -285.8 (to liquid water); Delta_vapH(H2O) = +44.0. For the water-gas shift CO(g) + H2O(g) -> CO2(g) + H2(g), select the correct statement(s).

Delta_fH deg(H2O(g)) = -241.8 kJ mol^-1

Delta_fH deg(CO) = -110.5 kJ mol^-1

Delta H for the shift reaction is -41.2 kJ

Delta_fH deg(H2O(g)) = -241.8 kJ mol^-1

Delta_fH deg(CO) = -110.5 kJ mol^-1

Delta H for the shift reaction is -41.2 kJ

The same reaction written with H2O(l) in place of H2O(g) has Delta H = -41.2 kJ

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3td092

Data in $kJ mo l^{- 1}$ : $Δ_{c} H (graphite) = - 393.5$ ; $Δ_{c} H (CO) = - 283.0$ ; $Δ_{c} H (H X_{2}) = - 285.8$ (to liquid water); $Δ_{v a p} H (H X_{2} O) = + 44.0$ . For the water-gas shift $CO (g) + H X_{2} O (g) CO X_{2} (g) + H X_{2} (g)$ , select the correct statement(s).

A $Δ_{f} H^{\circ} (H X_{2} O (g)) = - 241.8 kJ mo l^{- 1}$
B $Δ_{f} H^{\circ} (CO) = - 110.5 kJ mo l^{- 1}$
C $Δ H$ for the shift reaction is $- 41.2 kJ$
DThe same reaction written with $H X_{2} O (l)$ in place of $H X_{2} O (g)$ has $Δ H = - 41.2 kJ$

Δ_{f} H (CO) = - 393.5 - (- 283.0) = - 110.5

Δ_{f} H (H X_{2} O (g)) = - 285.8 + 44.0 = - 241.8

. Shift:

Δ H = - 393.5 - (- 110.5) - (- 241.8) = - 41.2 kJ

. With liquid water as reactant its formation enthalpy is more negative, so

Δ H = - 393.5 - (- 110.5) - (- 285.8) = + 2.8 kJ

, not

- 41.2 kJ