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 $ .



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