1) A reactions has a delta H of -76 Kj and a delta S of -117 J/K. Is the reaction spontaneous at 298 K?

Question

1) A reactions has a delta H of -76 Kj and a delta S of -117 J/K. Is the reaction spontaneous at 298 K?

Work:
Delta G= -76kJ/mol - 298K(-117J/K/1000)
Delta G= -41.134
Spontaneous

2) A reaction has a delta H of 11 kJ and a delta S of 49 J/K. Calculate the delta G at 298 K. Is the reaction spontaneous?

Work:
Delta G= 11kJ - 298K(49J/K/1000)
Delta G= -3.602
Spontaneous

Answer 1

Thanks for showing your work. These two look ok to me.

Answer 2

Thank you, I was just scared because they were back to back, and the only two problems of its kind, so I wasn't sure if it was right, or I made a mistake somewhere.

Answer 3

To determine if a reaction is spontaneous at a given temperature, we can use the equation:

ΔG = ΔH - TΔS

where:
ΔG = Gibbs free energy change
ΔH = enthalpy change
ΔS = entropy change
T = temperature in Kelvin

For the first reaction:
ΔH = -76 kJ
ΔS = -117 J/K
T = 298 K

ΔG = -76 kJ - 298 K(-117 J/K / 1000)
ΔG = -76 kJ + 298 K(-0.117 J/K)
ΔG = -76 kJ - 34.866 J
ΔG = -76 kJ - 0.034866 kJ
ΔG = -76.034866 kJ

Since ΔG is negative, the reaction is spontaneous at 298 K.

For the second reaction:
ΔH = 11 kJ
ΔS = 49 J/K
T = 298 K

ΔG = 11 kJ - 298 K(49 J/K / 1000)
ΔG = 11 kJ - 298 K(0.049 J/K)
ΔG = 11 kJ - 14.602 J
ΔG = 11 kJ - 0.014602 kJ
ΔG = 10.985398 kJ

Since ΔG is positive, the reaction is not spontaneous at 298 K.

Answer 4

To determine whether a reaction is spontaneous at a given temperature, we can use the Gibbs free energy equation:

ΔG = ΔH - TΔS

where ΔG is the change in Gibbs free energy, ΔH is the change in enthalpy, ΔS is the change in entropy, and T is the temperature in Kelvin.

1) For the first reaction with a ΔH of -76 kJ and a ΔS of -117 J/K, we can use the formula:

ΔG = -76 kJ - 298 K * (-117 J/K / 1000)

Here, we first convert the ΔS value from J/K to kJ/K by dividing by 1000. Then, we substitute the given values and calculate the result:

ΔG = -76 kJ - (-34.866 kJ)
ΔG = -76 kJ + 34.866 kJ
ΔG = -41.134 kJ

Since the resulting ΔG value is negative (-41.134 kJ), the reaction is spontaneous at 298 K.

2) For the second reaction with a ΔH of 11 kJ and a ΔS of 49 J/K, we can use the same formula:

ΔG = 11 kJ - 298 K * (49 J/K / 1000)

Again, we convert ΔS from J/K to kJ/K by dividing by 1000, and substitute the given values:

ΔG = 11 kJ - 298 K * (0.049 kJ/K)
ΔG = 11 kJ - 14.602 kJ
ΔG = -3.602 kJ

Once more, the resulting ΔG value is negative (-3.602 kJ), indicating that the reaction is spontaneous at 298 K.