Thermodynamics Question 101
Question: When $ 50cm^{3} $ of $ 0.2NH _{2}SO _{4} $ is mixed with $ 50cm^{3} $ of $ 1NKOH $ , the heat liberated is
[KCET 2004]
Options:
A) 11.46 kJ
B) 57.3 kJ
C) 573 kJ
D) 573 J
Show Answer
Answer:
Correct Answer: D
Solution:
For complete neutralization of strong acid and strong base energy released is $ 57.32KJ/mol $
No. of mole of $ H _{2}SO _{4}=\frac{0.2\times 50}{1000}={{10}^{-2}} $
No. of mole of $ KOH=\frac{1}{1000}\times 50=5\times {{10}^{-2}} $ So $ =57.32\times {{10}^{-2}}=0.5732KJ $
$ =573.2Joule $ .
BETA
