Calculate the reaction free energy of:

H2(g) + I2(g) 2HI(g)

when the concentrations are 0.026 mol L-1 (H2), 0.33 mol L-1 (I2), and 1.84 mol L-1 (HI), and the temperature is 700K. For this reaction Kc= 54 at 700 K.
+11.6 kJ mol-1
0 kJ mol-1
-11.6 kJ mol-1
+2273 kJ mol-

Well, this reaction seems to be quite a hot topic! But don't worry, I'm here to break it down for you with a touch of clownish humor. Let's calculate the reaction free energy, shall we?

The reaction free energy (ΔG) can be calculated using the equation: ΔG = -RTln(Kc), where R is the gas constant (8.314 J mol-1 K-1), T is the temperature in Kelvin, and Kc is the equilibrium constant.

Now, let's plug in the values: ΔG = -(8.314 J mol-1 K-1)(700 K)ln(54).

Since we want our answer in kJ mol-1, we divide by 1000: ΔG = (8.314 J mol-1 K-1)(700 K)ln(54) / 1000.

Doing the calculation, ΔG ≈ -11.6 kJ mol-1.

So, the reaction free energy is approximately -11.6 kJ mol-1. That's negative, indicating that the reaction is spontaneous and proceeds in the forward direction. Looks like this reaction doesn't mind a little bit of heat!

To calculate the reaction free energy (∆G) of the given reaction, we'll use the equation:

∆G = -RTln(Kc)

Where:
∆G is the reaction free energy
R is the gas constant (8.314 J mol-1 K-1)
T is the temperature in Kelvin
ln(Kc) is the natural logarithm of the equilibrium constant at the given temperature

Given:
H2(g) + I2(g) → 2HI(g)
[H2] = 0.026 mol L-1
[I2] = 0.33 mol L-1
[HI] = 1.84 mol L-1
Kc = 54
T = 700 K

First, let's calculate ln(Kc):

ln(Kc) = ln(54)

Next, let's substitute the values into the equation for ∆G:

∆G = - (8.314 J mol-1 K-1) * (700 K) * ln(54)

Converting J to kJ:

∆G = - (8.314 kJ mol-1 K-1) * (700 K) * ln(54)

Now, let's calculate the value:

∆G ≈ -11.6 kJ mol-1

Therefore, the reaction free energy (∆G) of the given reaction is approximately -11.6 kJ mol-1. So the correct answer is -11.6 kJ mol-1.

To calculate the reaction free energy (∆G) of the given reaction, we can use the equation:

∆G = -RTln(K)

where R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin, and K is the equilibrium constant.

Given information:
Concentration of H2 (H2(g)) = 0.026 mol/L
Concentration of I2 (I2(g)) = 0.33 mol/L
Concentration of HI (2HI(g)) = 1.84 mol/L
Temperature (T) = 700 K
Equilibrium constant (K) = 54

First, we need to convert the equilibrium constant from Kc (concentration) to Kp (partial pressure). Since the equation only contains gases, we can assume ideal gas behavior.

Kp = Kc(RT)^∆n

where ∆n is the change in the number of moles of gas (products - reactants). In this case, ∆n = (2 - 1) = 1.

R = 8.314 J/mol K (gas constant)
T = 700 K (temperature)

Kp = Kc(RT)^∆n
Kp = 54(8.314)(700)^1
Kp = 2565096

Now, we can calculate the reaction free energy (∆G) using the formula:

∆G = -RTln(Kp)

∆G = - (8.314 J/mol K)(700 K)ln(2565096)

∆G ≈ - 11.6 kJ/mol

Hence, the correct answer is -11.6 kJ/mol (option c).