Question

Calculate delta S surrounding when one mole of methanol (CH3OH) is formed from its elements under standard conditions if deltaf H° (CH3OH) =-238.9kJ/mol

Answer 1

∆S is the change in entropy . As ∆H is given and it is mentioned in the question that at standard contions temperature is 273 K then we can apply ∆H=T∆S

Answer 2

Therefore, Δ S (surrounding) = 801.6 $JK^-1mol^-1$

Explanation:

For the given reaction the thermodynamic equation can be represented as -

$CH_3OH (l) + \frac{3}{2} O_2 (g)$ → $CO_2 + 2H_2O (l)$

Given ;

Δ H° of methanol = -238.9 $kJmol^-1$

From the above equation, it is to be known that when one mole of methanol is formed from its constituent elements under standard conditions the heat released is of -238.9 $kJmol^-1$. Thus, the amount of energy released is absorbed by its surroundings.

∴ $q_s_u_r_r$. = +238.9 $kJmol^-1$

As we know that -

Δ S (surrounding) = $\frac{q_s_u_r_r.}{T}$

   = $\frac{238.9 kJmol^-1}{298 K}$

= 0.8016 $kJK^-1mol^-1$

= 801.6 $JK^-1mol^-1$

Therefore, Δ S (surrounding) = 801.6 $JK^-1mol^-1$