Question:
The $\Delta_{r}G$ of a reaction $A + B \rightleftharpoons C + D$ at fixed concentration of C is $-183.14 \, \text{kJ/mol}$. It is plotted as a function of temperature. Calculate the $G$ of the reaction (in kJ/mol).
The $\Delta_{r}G$ of a reaction $A + B \rightleftharpoons C + D$ at fixed concentration of C is $-183.14 \, \text{kJ/mol}$. It is plotted as a function of temperature. Calculate the $G$ of the reaction (in kJ/mol).
Show Hint
For $\Delta G$ vs $T$ graph: slope = $-\Delta S$ and intercept = $\Delta H$.
Updated On: Jul 18, 2026
- $-$330
- $+$330
- $-$349
- $+$349
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Write Gibbs free energy relation.
The variation of Gibbs free energy with temperature is: \[ \Delta G = \Delta H - T\Delta S \] This is a linear equation of form: \[ y = c + mx \] where intercept = $\Delta H$ and slope = $-\Delta S$.
Step 2: Read values from graph.
From graph: \[ T_1 = 200\,K,\quad \Delta G_1 = -183.14\,\text{kJ/mol} \] \[ T_2 = 300\,K,\quad \Delta G_2 = -100\,\text{kJ/mol} \]
Step 3: Calculate slope.
\[ m = \frac{-100 - (-183.14)}{300 - 200} = \frac{83.14}{100} = 0.8314 \] So, \[ \Delta S = -0.8314\,\text{kJ mol}^{-1}K^{-1} \]
Step 4: Find intercept.
\[ \Delta G = 0.8314T + b \] \[ -183.14 = 0.8314(200) + b \] \[ b = -349.42 \approx -349 \]
Step 5: Interpret result.
The intercept corresponds to standard Gibbs free energy change $\Delta G^\circ$.
Step 6: Final answer.
\[ \boxed{-349 \, \text{kJ/mol}} \]
The variation of Gibbs free energy with temperature is: \[ \Delta G = \Delta H - T\Delta S \] This is a linear equation of form: \[ y = c + mx \] where intercept = $\Delta H$ and slope = $-\Delta S$.
Step 2: Read values from graph.
From graph: \[ T_1 = 200\,K,\quad \Delta G_1 = -183.14\,\text{kJ/mol} \] \[ T_2 = 300\,K,\quad \Delta G_2 = -100\,\text{kJ/mol} \]
Step 3: Calculate slope.
\[ m = \frac{-100 - (-183.14)}{300 - 200} = \frac{83.14}{100} = 0.8314 \] So, \[ \Delta S = -0.8314\,\text{kJ mol}^{-1}K^{-1} \]
Step 4: Find intercept.
\[ \Delta G = 0.8314T + b \] \[ -183.14 = 0.8314(200) + b \] \[ b = -349.42 \approx -349 \]
Step 5: Interpret result.
The intercept corresponds to standard Gibbs free energy change $\Delta G^\circ$.
Step 6: Final answer.
\[ \boxed{-349 \, \text{kJ/mol}} \]
Was this answer helpful?
0
0
Top AP EAPCET Chemistry Questions
- The four quantum numbers of the valence electron of potassium are :
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
- In which of the following the election gain enthalpy of elements is correctly arranged?
- The radii of $ \text{F,}{{\text{F}}^{-}}\text{O} $ and $ {{\text{O}}^{2-}} $ are in the order
- Which of the following has the highest first ionization potential?
View More Questions
Top AP EAPCET Thermodynamics Questions
- For one mole of an ideal gas, three isochors were obtained at \(V_1, V_2\) and \(V_3\) respectively. Their slopes are \(m_1, m_2\) and \(m_3\). If \(V_1<V_2<V_3\), then the correct relationship of slopes is:
- The enthalpy change (ΔH) for the reaction N2(g) + 3H2(g) → 2NH3(g) is -92 kJ/mol.
What is the enthalpy change for the decomposition of 1 mol of NH3(g) into its elements? - Identify the correct statements from the following. (i). For adiabatic process, $\Delta U = W_{adiabatic$} (ii). Work is a path function. (iii). Volume is an extensive property.
- Hydration enthalpy of which of the following ions is highest?
- The standard Gibbs energy ($\Delta G^\ominus$) for the following reaction is A(s) + B$^{2+}$ (aq) $\rightleftharpoons$ A$^{2+}$ (aq) + B(s), K$_c$=10$^{12}$ at 25$^\circ$C (K$_c$= equilibrium constant)
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions