Δ H°f (298 K) of methanol is given by the chemical equation

Question

Δ H°f (298 K) of methanol is given by the chemical equation (A) CH4(g) + ( 1/2 )O2(g) → CH3OH(g) (B) C (graphite) + ( 1/2 )O2(g) + 2H2(g) → CH3OH(l) (C) C (diamond) + ( 1/2 )O2(g) + 2H2(g) → CH3OH(l) (D) CO(g) +2H2(g) → CH3OH(l)

Tardigrade Admin · Accepted Answer

Heat of formation is defined as the heat exchange when one mole of a substance is formed from its constituent elements at STP. In case of allotropes most stable allotrope is taken into account. Among graphite and diamond, graphite is stabler. Hence, heat of formation of

C H_{3} O H

is represented by equation.

C

(graphite)

+ \frac{1}{2} O_{2} (g) + 2 H_{2} (g) ⟶ C H_{3} O H (l)

Q. $Δ H_{f}^{\circ} (298 K)$ of methanol is given by the chemical equation

$C H_{4 (g)} + \frac{1}{2} O_{2 (g)} \to C H_{3} O H_{(g)}$

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

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

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

Q. ΔHf∘​(298K) of methanol is given by the chemical equation

CH4(g)​+21​O2(g)​→CH3​OH(g)​

C (graphite) +21​O2(g)​+2H2(g)​→CH3​OH(l)​

C (diamond) +21​O2(g)​+2H2(g)​→CH3​OH(l)​

CO(g)​+2H2(g)​→CH3​OH(l)​

Q. $Δ H_{f}^{\circ} (298 K)$ of methanol is given by the chemical equation

$C H_{4 (g)} + \frac{1}{2} O_{2 (g)} \to C H_{3} O H_{(g)}$

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

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

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