NEET Chemistry Chemical Equilibrium MCQ Question
Calculate the change in Gibbs free energy (∆G ) for the formation of NO2 from NO and O2 at 298K using the given standard Gibbs free energy of formation values: ∆fG (NO2) = 52.0 kJ/mol, ∆fG (NO) = 87.0 kJ/mol, ∆fG (O2) = 0 kJ/mol.
-35.0 kJ/mol
-87.0 kJ/mol
35.0 kJ/mol
87.0 kJ/mol
Correct Answer
Detailed Explanation
The change in Gibbs free energy (∆G ) is calculated using the formula: ∆G = ∆fG (products) - ∆fG (reactants). Substituting the given values: ∆G = 52.0 - (87.0 + 0) = -35.0 kJ/mol.
Found an issue with this question?
Related Questions
More from Chemical Equilibrium
In the diagram shown above, which graph best represents the change in concentration of H2 as the reaction H2(g) + I2(g) → 2HI(g) proceeds to equilibri...
Match Column-I with Column-II. Column-I Column-II (a) K c for H 2 + I 2 2HI (i) 57.0 (b) K c for N 2 + O 2 2NO (ii) 4.8 × 10 –31 (c) K...
More from
Assertion (A): For a dibasic acid H2X, the equilibrium constant for the first ionization step (Ka1) is generally greater than that for the second ioni...
Assertion (A): The solubility of a sparingly soluble salt is affected by the presence of a common ion. | Reason (R): The solubility product constant (...
Match Column-I with Column-II. Column-I Column-II (a) Solubility Product Constant (K sp) (i) Represents maximum concentration of ions in s...