N2(g)+3F2 yields 2Nf3(g)

Delta H(degree)298=-264KJ mol-1;Delta S(degree)298=-278J K-1 mol-1
The following questions relate to the synthesis reaction represented by the chemical equation above.

a.)CAlculate the value of the standard free energy change, Delta G(degree)298, for the reaction.

b.) Determine the temperature at which the equilibrium constant, Keq, for the reaction is equal to 1.00. (Assume that Delta H and Delta S are independent of temperature)

c.)Calculate the stnadard enthalpy change, Delta H, that occurs when a 0.256 mol sample of NF3(g), is formed from N2 and F2 at 1.00atm and 298K.

a) To calculate the standard free energy change (ΔG°298) for the reaction, you can use the equation:

ΔG°298 = ΔH°298 - TΔS°298

where ΔH°298 is the standard enthalpy change and ΔS°298 is the standard entropy change at 298 K. T represents the temperature in Kelvin.

Given:
ΔH°298 = -264 kJ/mol
ΔS°298 = -278 J/(K*mol)

First, convert the units of ΔS°298 to kJ/(K*mol) for the equation:

ΔS°298 = -278 J/(K*mol) * (1 kJ/1000 J)
ΔS°298 = -0.278 kJ/(K*mol)

Now substitute the values into the equation:

ΔG°298 = -264 kJ/mol - T * (-0.278 kJ/(K*mol))

b) To determine the temperature at which the equilibrium constant (Keq) for the reaction is equal to 1.00, you can use the equation:

ΔG° = -RT ln(Keq)

where ΔG° is the standard free energy change, R is the gas constant (8.314 J/(mol*K)), and ln represents the natural logarithm.

Given that Keq = 1.00, we can rearrange the equation as follows:

1.00 = exp(-ΔG°/(RT))

Take the natural logarithm of both sides:

ln(1.00) = -ΔG°/(RT)

Since ln(1.00) is equal to 0, the equation simplifies to:

0 = -ΔG°/(RT)

Now we can solve for the temperature (T):

T = -ΔG°/(R * 0)
T = 0 K

Therefore, the equilibrium constant (Keq) is equal to 1.00 at 0 Kelvin.

c) To calculate the standard enthalpy change (ΔH) when a 0.256 mol sample of NF3(g) is formed from N2 and F2 at 1.00 atm and 298 K, you need to use the equation:

ΔH = ΔH° + ΣνΔH°f(products) - ΣνΔH°f(reactants)

where ΔH is the enthalpy change, ΔH° is the standard enthalpy change, ν is the stoichiometric coefficient, and ΔH°f is the standard enthalpy of formation.

In this case, you have:

Reactants: N2(g) and 3F2(g)
Products: 2NF3(g)

Given the stoichiometry, we can rewrite the equation as:

ΔH = ΔH°f(NF3(g)) - ΔH°f(N2(g)) - 3ΔH°f(F2(g))

Now, you need to look up the standard enthalpy of formation values for each species involved and substitute them into the equation.