Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Estimate the temperature range for which the following standard reaction is product-favored:

$\mathrm{ 2 \mathrm{H}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{H}_2 \mathrm{O}(g) }$

Given that the standard enthalpy change $\mathrm{ \left(\Delta H^{\circ}\right) }$ for the reaction is $\mathrm{-483.6 \: \mathrm{kJ} / \mathrm{mol}}$ and the standard entropy change $\mathrm{\left(\Delta S^{\circ}\right)\: is \: 140.3 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}).}$

Option: 1
Product favoured above $-3448 \mathrm{~K}$

Option: 2
Not product favoured above $-3448 \mathrm{~K}$

Option: 3
Product favoured above $-3400 \mathrm{~K}$

Option: 4
Not Product favoured $2300 \mathrm{~K}$

Answers (1)

best_answer

Step 1: Calculate the standard Gibbs free energy change $\mathrm{\left(\Delta G^{\circ}\right)}$ using the equation:
$\mathrm{ \Delta G^{\circ}=\Delta H^{\circ}-T \cdot \Delta S^{\circ} }$

Step 2: Determine the temperature range for which the reaction is productfavored. In this case, a reaction is product-favored when $\mathrm{ \Delta G^{\circ} }$ is negative.

Step 3: Set $\mathrm{ \Delta G^{\circ} }$ to be less than zero to find the temperature range:

$\mathrm{ \Delta G^{\circ}<0 \Rightarrow \Delta H^{\circ}-T \cdot \Delta S^{\circ}<0 }$

Step 4: Solve for the temperature $\mathrm{ (T) }$ to find the range where the reaction is product-favored.
$\mathrm{ T>\frac{\Delta H^{\circ}}{\Delta S^{\circ}} }$

Step 5: Substitute the given values and calculate the temperature range.
$\mathrm{ T>\frac{-483.6 \times 10^3 \mathrm{~J} / \mathrm{mol}}{140.3 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})} }$

Calculating the numerical value of the temperature range:

$\mathrm{ T>-3447.76 \mathrm{~K} }$

Since temperature cannot be negative, the reaction is product-favored for all temperatures above $\mathrm{ -3447.76 \mathrm{~K}. }$

So, Correct option is 1

Posted by

SANGALDEEP SINGH

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pK_b of ammonia solution is 4.75, the pH of the mixture will be :
Option: 1 3.75
Option: 2 4.75

Trending Articles/News

anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

anam-khan

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs

anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question