Question

Calculate ∆Ssurr when one mole of methanol
(CH3OH) is formed form its elements under
standard conditions if
∆fH°CH OH) =-238.9 kJ mol-1
​

Answer 1

Answer:

Delta S = -Delta H/temperature

Explanation:

S = 238.9/25 = 9.556 kJ mol-1 k-1

Answer 2

Given: ΔH° of CH3OH = -238.9 KJ mol^(-1)

Temperature = 25°C

To Find:- ΔS (surrounding) =?

Solution:-

Second law of thermodynamics states that the entropy can in no way decrease over time in an isolated system and it shall remain consistent type all of the processes conducted are reversible.

- According to second law of thermodynamics ΔS°(surr) must be positive because ΔS°(sys) is negative.

- ΔS°(uni)  ≥ 0 for reverssible then only

ΔS°(surr) >0.

- It is given that ΔH° of methanol is

= -238.9 KJ mol^(-1)

= -238900 Jmol^(-1)

and, temperature under standard condition is 25°C

i.e 25 + 273 = 298 K

- System will loses heat to surrounding as the reaction is exothermic due to negative sign of ΔH°

- Hence, The enthalpy of surrounding increases,

ΔH°(surr) = + 238900

- As we know that,

ΔS°(surr) = ΔH°(surr)/T

ΔS°(surr) = 238900/298

= 801.67 Jmol^(-1)K^(-1)

- Hence, The ΔS°(surr) of methanol is

801.67 Jmol^(-1)K^(-1)