Given: C(s) + O₂(g) → CO₂(g), Δ H₁ = -393.5 kJ/mol H₂(g) + (1)/(2)O₂(g) → H₂O(l), Δ H₂ = -285.8 kJ/mol C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), Δ H₃ = -1366.5 kJ/mol Find Δ Hf of ethanol.

Target: 2C + 3H₂ + (1)/(2)O₂ → C₂H₅OH Δ Hf = 2Δ H₁ + 3Δ H₂ - Δ H₃ = 2(-393.5) + 3(-285.8) - (-1366.5) = -787 - 857.4 + 1366.5 = -277.9 kJ/mol

JEE chemistry

Given: C(s) + O₂(g) → CO₂(g), Δ H₁ = -393.5 kJ/mol H₂(g) + (1)/(2)O₂(g) → H₂O(l), Δ H₂ = -285.8 kJ/mol C₂H₅OH(l) + — JEE Chemistry

PPPriya Patel · 12 Asked 20d ago 1,248 views 1 answer

Given:

$C(s) + O_2(g) \rightarrow CO_2(g)$, $\Delta H_1 = -393.5$ kJ/mol
$H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)$, $\Delta H_2 = -285.8$ kJ/mol
$C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$, $\Delta H_3 = -1366.5$ kJ/mol

Find $\Delta H_f$ of ethanol.

KKavyaSharma27 ✓ Accepted · 19d ago ▲ 33

Target: $2C + 3H_2 + \frac{1}{2}O_2 \rightarrow C_2H_5OH$

$$\Delta H_f = 2\Delta H_1 + 3\Delta H_2 - \Delta H_3$$

$$= 2(-393.5) + 3(-285.8) - (-1366.5)$$

$$= -787 - 857.4 + 1366.5 = -277.9 kJ/mol$$

Does this approach generalise to the JEE Advanced version of this question?

RiteshBasnet93 · 18d ago

Is there a faster shortcut for this in the actual exam? Time is tight.

Nikhil Rao · 16d ago

Why do we take the positive value only in the last step?

PrakashGurung52 · 16d ago