Question

C2H4(g) + H2(g) → C2H6(g) ΔH = –137.5 kJ; ΔS = –120.5 J/K Calculate ΔG at 25 °C and determine whether the reaction is spontaneous. Does ΔG become more negative or more positive as the temperature increases?

Answer 1

Answer:-ΔG=-101.5KJ

Explanation:We have to calculate ΔG for the reaction  so using the formula given in the equation we can calculate the \Delta G for the reaction.

We need to convert the unit ofΔS in terms of KJ/Kelvin as its value is given in terms of J/Kelvin

Also we need to convert the temperature in Kelvin as it is given in degree celsius.

$\Delta H=-137.5\\ \Delta S=-120J/K\\ \Delta S=-0.120KJ/K\\ T=25^{.C}\\ T=273+25=298 K\\ \Delta G=?\\ \Delta G=\Delta H-T\Delta S\\\Delta G=-137.5KJ-(278\times -0.120)\\ \Delta G=-137.5+35.76\\\Delta G=-101.74\\\Delta G=Negative$

After calculating forΔG we found that the value ofΔG is negative and its value is -101.74KJ

 For a  reaction to be spontaneous the value of \Delta G \ must be negative .

As the ΔG for the given reaction is is negative so the reaction will be spontaneous in nature.

In this reaction since the entropy of reaction is positive and hence when we increase the temperature term then the overall term TΔS would become more positive  and hence the value of ΔG would be less negative .

Hence the value of ΔG  would become more positive with the increase in temperature.

So we found the value of ΔG to be -101.74KJ

Answer 2

Answer:

ΔG = -101.591 KJ

Explanation:

Gibbs free energy -

It is a thermodynamic quantity , which is given by the change in enthalpy minus the product of the change in entropy and absolute temperature.

i.e.,

ΔG is given as the change in gibbs free energy ( KJ )

ΔS is given as the change in entropy ( KJ /K )

ΔH is given as the change in ethalphy ( KJ )

T = temperature ( Kelvin ( K ))

ΔG  =  ΔH - TΔS

The sign of ΔG determines the reaction spontaneity , as

ΔG = negative , the reaction is spontaneous and

If ΔG = positive , the reaction is non spontaneous .

Given -

For the reaction ,

C₂H₄ (g) + H₂(g) ---> C₂H₆(g)

ΔH = - 137.5 KJ

ΔS = - 120.5 J /K

Since ,

1 KJ = 1000 J

1 J = 1 / 1000KJ

ΔS = - 120.5 / 1000 KJ /K

ΔS = -0.1205 KJ /K

T = 25°C

(adding 273 To °C to convert it to K)

T = 25 + 273 = 298 K

Putting the values on the above equation ,

ΔG = ΔH - TΔS

ΔG = -137.5 KJ - 298 * (-0.1205 KJ / K)

ΔG = -137.5 KJ + 35.909 KJ

ΔG = -101.591 KJ

Since,

the value of ΔG is negative ,

hence, the reaction is spontaneous.

For the above reaction ,

If the temperature is increased ,

ΔG = ΔH - TΔS

From the above equation ,

the value of TΔS will increase ,

As a result the value of ΔG will be more positive , by increasing the temperature.