Question

The ∆G° of vaporization for butane at 298 K and 1.00 atm is -2.125 kJ/mol. Calculate the pressure, in atm, of butane vapor in equilibrium with butane liquid at 298 K.

Answer 1

Explanation:

X-axis=(5,0)   Y-axis =(0,5)

Comparing with (n,0)  and  (0,m)

n=5  and m=5

∴m+n=5+5=10

Answer 2

Answer:

1.78 atm is the correct answer.

Explanation:

The pressure of butane vapour in equilibrium with butane liquid at 298 K can be calculated using the following equation:

ΔG° = -RTln(K)

where ΔG° is the standard Gibbs free energy change of vaporization, R is the gas constant (8.314 J/(mol·K)), T is the temperature (298 K), and K is the equilibrium constant for the vaporization reaction.

To calculate the equilibrium pressure, we need to first calculate the equilibrium constant K:

K = e^(-ΔG°/RT)

Substituting the given values, we get:

K= $e^(-(-2.125 × 10^3 J/mol)/(8.314 J/(mol·K) × 298 K)) = 1.78$

The equilibrium pressure, P, can then be calculated from the equation:

$P = K * P°$

where P° is the standard pressure (1 atm).

Substituting the calculated value of K and P°, we get:

$P = 1.78 * 1 atm = 1.78 atm$

Therefore, the pressure of butane vapour in equilibrium with butane liquid at 298 K is 1.78 atm.

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