Calculate delta gº for the following system:

delta h = -472 kj/mol
delta s = -108 j/molk

answer is supposed to be 440 kj, but i have no clue how to get there

ΔG = - 442.5  KJ/mol    

Explanation:

Data Given

delta H = -472 kJ/mol

delta S = -108 J/mol K

So,

delta S = -0.108 J/mol K

delta Gº = ?

Solution:

The answer will be calculated by the following equation for the Gibbs free energy

                    G = H - TS

Where

G = Gibbs free energy

H = enthalpy of a system (heat

T = temperature

S = entropy

So the change in the Gibbs free energy at constant temperature can be written as

                ΔG = ΔH - TΔS . . . . . . (1)

Where

ΔG = Change in Gibb’s free energy

ΔH = Change in enthalpy of a system

ΔS = Change in entropy

if system have standard temperature then

T = 273.15 K

Now,

put values in equation 1

                  ΔG = (-472 kJ/mol) - 273.15 K (-0.108 KJ/mol K)

                 ΔG = (-472 kJ/mol) - (-29.5 KJ/mol)

                 ΔG = -472 kJ/mol + 29.5 KJ/mol

                 ΔG = - 442.5  KJ/mol          

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

b. th3 n4zc4 pl4t3 15 c0nv3rg1ng w1th the s0uth 4m3ric4n pl4t3.