Numerical/Formula

1. TITLE

How to solve Gibbs Energy & Equilibrium Constant numericals in NEET — Step-by-step

2. PROBLEM TYPE INTRO

This category of numerical connects Chemical Thermodynamics with Chemical Equilibrium. To solve these, you will need the standard Gibbs free energy formula relating to the equilibrium constant ($K$): $\Delta_r G^\ominus = -2.303RT \log K$. The most typical mistake students make is forgetting to convert standard Gibbs energy from kilojoules ($\text{kJ}$) to Joules ($\text{J}$) before plugging it into the equation alongside the gas constant ($R$).

3. MCQ

Question:
Find out the value of $\log K$ for the following reaction at $298 \text{ K}$:
$2NH_3(g) + CO_2(g) \rightleftharpoons NH_2CONH_2(aq) + H_2O(l)$
(Given: Standard Gibbs energy change, $\Delta_r G^\ominus$ is $-13.6 \text{ kJ mol}^{-1}$ and $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$)

(a) $2.38$
(b) $-2.38$
(c) $0.00238$
(d) $13.6$

4. STEP-BY-STEP SOLUTION

SOLUTION

Given:
* $\Delta_r G^\ominus = -13.6 \text{ kJ mol}^{-1}$
* $T = 298 \text{ K}$
* $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$

To Find:
The value of $\log K$.

Formula:
$\Delta_r G^\ominus = -2.303RT \log K$
Use this formula when relating the standard Gibbs energy change of a reaction directly to its equilibrium constant at a specific constant temperature.

Step 1: Convert units of $\Delta_r G^\ominus$
Because the universal gas constant ($R$) is given in Joules ($\text{J}$), you must convert the Gibbs energy from $\text{kJ}$ to $\text{J}$.
$\Delta_r G^\ominus = -13.6 \times 10^3 \text{ J mol}^{-1} = -13600 \text{ J mol}^{-1}$

Step 2: Rearrange the formula to solve for $\log K$
$\log K = \frac{-\Delta_r G^\ominus}{2.303 RT}$

Step 3: Substitute the values and calculate
$\log K = \frac{-(-13600 \text{ J mol}^{-1})}{2.303 \times (8.314 \text{ J K}^{-1} \text{mol}^{-1}) \times 298 \text{ K}}$
$\log K = \frac{13600}{5705.85}$
$\log K = 2.38$

(Note: To find the actual equilibrium constant $K$, you would take the antilog: $K = \text{antilog } 2.38 = 2.4 \times 10^2$)

Answer: (a) $2.38$

5. WHERE STUDENTS GO WRONG

  • ❌ Mistake 1: Forgetting to convert $\text{kJ}$ to $\text{J}$ (Option c). $R$ is given in $\text{J K}^{-1} \text{mol}^{-1}$, so $\Delta_r G^\ominus$ must be multiplied by $10^3$ to avoid a massive magnitude error.
  • ❌ Mistake 2: Missing the negative sign in the formula (Option b). The formula is $\Delta_r G^\ominus = \mathbf{-}2.303RT \log K$. Dropping the negative sign results in calculating an incorrect, negative $\log K$.
  • ❌ Mistake 3: Calculating $\ln K$ instead of $\log K$. If you forget to include the $2.303$ conversion factor, you will accidentally calculate the natural log ($\ln$) rather than the base-10 log.

6. QUICK FORMULA REFERENCE

Formulas Used:
* $\Delta_r G^\ominus = -2.303 RT \log K$ — Use when calculating base-10 logarithms to find the equilibrium constant, which is standard for most NEET numericals.
* $\Delta_r G^\ominus = -RT \ln K$ — Use when working directly with natural logarithms.
* $\Delta G = \Delta H - T\Delta S$ — Use to find Gibbs energy if enthalpy ($\Delta H$) and entropy ($\Delta S$) changes are provided instead of $K$.

7. SIMILAR PROBLEM HINT

"If NEET changes the values, the method stays the same. Practice with: calculating the exact value of $K$ using antilogs when $\Delta_r G^\ominus$ is a large negative value to see how it strongly favors product formation!"

8. NCERT REFERENCE

Based on: NCERT Class 11, Chapter Thermodynamics, Problem 5.13

9. CTA

More solved numericals → Chemical Thermodynamics Complete Guide
Practice problems → Mock exams