To calculate the Gibbs free energy (∆G), we can use the formula:

∆G = ∆H - T∆S

where:
∆H is the change in enthalpy (heat content),
T is the absolute temperature in Kelvin, and
∆S is the change in entropy (disorder).

Given:
∆H298K = 8 Kcal
T = 298 K
So for C(s) = 1.4 cal.deg-1mole-1
So for H2(g) = 31.2 cal.deg-1mole-1
So for C2H4(g) = 52.2 cal.deg-1mole-1

First, we need to calculate the change in entropy (∆S) for the reaction.

∆S = So (products) - So (reactants)

So, ∆S = So (C2H4) - [2*So (C) + 2*So (H2)]

Substitute the given values:

∆S = 52.2 - [2*1.4 + 2*31.2] = 52.2 - [2.8 + 62.4] = 52.2 - 65.2 = -13 cal.deg-1mole-1

Now, we can calculate ∆G using the formula:

∆G = ∆H - T∆S

But before that, we need to convert ∆S from cal.deg-1mole-1 to Kcal.deg-1mole-1 by dividing it by 1000 (since 1 Kcal = 1000 cal). So, ∆S = -13/1000 = -0.013 Kcal.deg-1mole-1

Now, substitute the values into the formula:

∆G = 8 - 298*(-0.013) = 8 - (-3.874) = 8 + 3.874 = 11.874 Kcal

Since ∆G is positive, the reaction is not spontaneous at 298K. The reaction would be spontaneous if ∆G were negative.

Question

To calculate the Gibbs free energy (∆G), we can use the formula:

∆G = ∆H - T∆S

where:
∆H is the change in enthalpy (heat content), 
T is the absolute temperature in Kelvin, and 
∆S is the change in entropy (disorder).

Given:
∆H298K = 8 Kcal
T = 298 K
So for C(s) = 1.4 cal.deg-1mole-1
So for H2(g) = 31.2 cal.deg-1mole-1
So for C2H4(g) = 52.2 cal.deg-1mole-1

First, we need to calculate the change in entropy (∆S) for the reaction.

∆S = So (products) - So (reactants)

So, ∆S = So (C2H4) - [2*So (C) + 2*So (H2)]

Substitute the given values:

∆S = 52.2 - [2*1.4 + 2*31.2] = 52.2 - [2.8 + 62.4] = 52.2 - 65.2 = -13 cal.deg-1mole-1

Now, we can calculate ∆G using the formula:

∆G = ∆H - T∆S

But before that, we need to convert ∆S from cal.deg-1mole-1 to Kcal.deg-1mole-1 by dividing it by 1000 (since 1 Kcal = 1000 cal). So, ∆S = -13/1000 = -0.013 Kcal.deg-1mole-1

Now, substitute the values into the formula:

∆G = 8 - 298*(-0.013) = 8 - (-3.874) = 8 + 3.874 = 11.874 Kcal

Since ∆G is positive, the reaction is not spontaneous at 298K. The reaction would be spontaneous if ∆G were negative.

Knowee AI · Accepted Answer

To calculate the Gibbs free energy (∆G), we can use the formula:

∆G = ∆H - T∆S

where:
∆H is the change in enthalpy (heat content), 
T is the absolute temperature in Kelvin, and 
∆S is the change in entropy (disorder).

Given:
∆H298K = 8 Kcal
T = 298 K
So for C(s) = 1.4 cal.deg-1mole-1
So for H2(g) = 31.2 cal.deg-1mole-1
So for C2H4(g) = 52.2 cal.deg-1mole-1

First, we need to calculate the change in entropy (∆S) for the reaction.

∆S = So (products) - So (reactants)

So, ∆S = So (C2H4) - [2*So (C) + 2*So (H2)]

Substitute the given values:

∆S = 52.2 - [2*1.4 + 2*31.2] = 52.2 - [2.8 + 62.4] = 52.2 - 65.2 = -13 cal.deg-1mole-1

Now, we can calculate ∆G using the formula:

∆G = ∆H - T∆S

But before that, we need to convert ∆S from cal.deg-1mole-1 to Kcal.deg-1mole-1 by dividing it by 1000 (since 1 Kcal = 1000 cal). So, ∆S = -13/1000 = -0.013 Kcal.deg-1mole-1

Now, substitute the values into the formula:

∆G = 8 - 298*(-0.013) = 8 - (-3.874) = 8 + 3.874 = 11.874 Kcal

Since ∆G is positive, the reaction is not spontaneous at 298K. The reaction would be spontaneous if ∆G were negative.

Calculate the Gibbs free energy for formation of C2H4. Comment on the spontaneityor otherwise of the reaction at 298K.Given 2C(s)+ 2H2(g)→C2H4(g); ∆H298K = 8Kcal.Substance So(in cal.deg-1mole-1)C(s) 1.4H2(g) 31.2C2H4(g) 52.2

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