Use the table below to answer the questions that follow.
Thermodynamic Quantities for Selected Substances at 298.15 K (25°C)
-Consider the reaction:

Given the following table of thermodynamic data at 298 ^oK:

The value of K for the reaction at 25 °C is .

Question

Use the table below to answer the questions that follow.

Thermodynamic Quantities for Selected Substances at 298.15 K (25°C) Use the table below to answer the questions that follow. Thermodynamic Quantities for Selected Substances at 298.15 K (25°C) -Consider the reaction: Given the following table of thermodynamic data at 298 oK: The value of K for the reaction at 25 °C is . A) 1.8 × 104 B) 1.9 × 10- 10 C) 5.4 × 109 D) 810 E) 3.7 × 1010

-Consider the reaction:
Use the table below to answer the questions that follow. Thermodynamic Quantities for Selected Substances at 298.15 K (25°C) -Consider the reaction: Given the following table of thermodynamic data at 298 oK: The value of K for the reaction at 25 °C is . A) 1.8 × 104 B) 1.9 × 10- 10 C) 5.4 × 109 D) 810 E) 3.7 × 1010

Given the following table of thermodynamic data at 298 ^oK:
Use the table below to answer the questions that follow. Thermodynamic Quantities for Selected Substances at 298.15 K (25°C) -Consider the reaction: Given the following table of thermodynamic data at 298 oK: The value of K for the reaction at 25 °C is . A) 1.8 × 104 B) 1.9 × 10- 10 C) 5.4 × 109 D) 810 E) 3.7 × 1010

The value of K for the reaction at 25 °C is .

Quizplus · Accepted Answer

We can use the equation ΔG° = -RT ln K to solve for K. From the table, we have:
ΔG° = (-394.4 kJ/mol) - 2(-285.8 kJ/mol) + 4(213.7 kJ/mol) = -1637.6 kJ/mol
Converting to J/mol and plugging in the values (R = 8.314 J/mol-K, T = 298 K), we get:
-1637600 J/mol = -8.314 J/mol-K * 298 K * ln K
Solving for K, we get:
K = 5.4 × 10^19

Use the Table Below to Answer the Questions That Follow