Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The standard enthalpy change <img alt="\mathrm{\left(\Delta H^{\circ}\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cleft%28%5CDelta%20H%5E%7B%5Ccirc%7D%5Cright%29%7D"

The standard enthalpy change \mathrm{\left(\Delta H^{\circ}\right)} for a reaction is \mathrm{-500 \mathrm{~kJ} / \mathrm{mol}}. Calculate the heat (q) released when 2 moles of the reaction take place at constant pressure.

Option: 1

1000 kJ


Option: 2

500 kJ


Option: 3

-500 kJ


Option: 4

-1000 kJ


Answers (1)

best_answer

The heat (q) released during a reaction at constant pressure is equal to the negative of the enthalpy change \mathrm{\left(\Delta H^{\circ}\right)} multiplied by the number of moles (n) of the reaction:

\mathrm{ q=-\Delta H^{\circ} \cdot n=-(-500 \mathrm{~kJ} / \mathrm{mol}) \cdot 2 \mathrm{~mol}=1000 \mathrm{~kJ} }

So, the correct answer is: A) 1000 \mathrm{~kJ}

Posted by

Sayak

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

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