Which of the following equations corresponds to the definition of enthalpy of formation at 298 K ?

Question

Which of the following equations corresponds to the definition of enthalpy of formation at 298 K ? (A) C (graphite) +2 H 2( g )+1 / 2 O 2(l) arrow CH 3 OH ( g ) (B) C (diamond) +2 H 2( g )+1 / 2 O 2( g ) arrow CH 3 OH (l) (C) 2 C (graphite) +4 H 2( g )+ O 2( g ) arrow 2 CH 3 OH (l) (D) C (graphite) +2 H 2( g )+1 / 2 O 2( g ) arrow CH 3 OH (l)

Tardigrade Admin · Accepted Answer

In enthalpy of formation, reactants and products must be in most stable standard state

Q. Which of the following equations corresponds to the definition of enthalpy of formation at $298 K$ ?

$C$ (graphite) $+ 2 H_{2} (g) + 1/2 O_{2} (l) \to C H_{3} O H (g)$

$C$ (diamond) $+ 2 H_{2} (g) + 1/2 O_{2} (g) \to C H_{3} O H (l)$

$2 C$ (graphite) $+ 4 H_{2} (g) + O_{2} (g) \to 2 C H_{3} O H (l)$

$C$ (graphite) $+ 2 H_{2} (g) + 1/2 O_{2} (g) \to C H_{3} O H (l)$

In enthalpy of formation, reactants and products must be in most stable standard state

Q. Which of the following equations corresponds to the definition of enthalpy of formation at 298K ?

C (graphite) +2H2​(g)+1/2O2​(l)→CH3​OH(g)

C (diamond) +2H2​(g)+1/2O2​(g)→CH3​OH(l)

2C (graphite) +4H2​(g)+O2​(g)→2CH3​OH(l)

C (graphite) +2H2​(g)+1/2O2​(g)→CH3​OH(l)

In enthalpy of formation, reactants and products must be in most stable standard state

Q. Which of the following equations corresponds to the definition of enthalpy of formation at $298 K$ ?

$C$ (graphite) $+ 2 H_{2} (g) + 1/2 O_{2} (l) \to C H_{3} O H (g)$

$C$ (diamond) $+ 2 H_{2} (g) + 1/2 O_{2} (g) \to C H_{3} O H (l)$

$2 C$ (graphite) $+ 4 H_{2} (g) + O_{2} (g) \to 2 C H_{3} O H (l)$

$C$ (graphite) $+ 2 H_{2} (g) + 1/2 O_{2} (g) \to C H_{3} O H (l)$