When `100` ml of `0.2 M HCl` is mixed with 100 ml of `0.2 M NaOH`, the rise in temperature is `T_(1)`. When the experiment is repeated using 200 ml each of the same solutions, the rise in temperature is `T_(2)`. Then

Options

A. `T_(1)= T_(2)`
B. `T_(2)= 2T_(2)`
C. `T_(1)= 4_(2)`
D. `T_(2)= 9T_(1)`

Answer

A

Explanation

No. of moles of KOH `=0.2 xx 0.5= 0.1` moles
No. of moles of HCl `=0.2 xx 0.5= 0.1` moles
Heat evolved due to neutralization `=Q_(1)`
When the experiment is repeated using half of the solution then the no. of moles of both the solution will be reduced to half of the original. Thus heat evolved `Q_(2)= Q_(1//2)`
But in second case the volume of solution (250mL + 250mL) also reduced to half of the original (500mL + 500mL). So the increase in temperature will be same i.e., `T_(1)= T_(2)`