Question

100 ml of 0.2 m h2so4 is reacted with 100 ml of 0.5m naoh solution .the heat released in the neutralisation process

Answer 1

The heat released in the neutralisation process is -2.3 kJ

Explanation:

Heat of neutralization defined as heat evolved when one gram equivalent of an acid is completely neutralized by one gram equivalent of a base.

Therefore, the neutralization between strong acid and strong base may also be called as formation of one mole of water.

So, the heat change accompanying this reaction is constant and the value is -57 kJ/mol

To calculate the number of moles for given molarity, we use the equation:

$\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}$  

a) Molarity of $H_2SO_4$ solution = 0.2 M

Volume of $H_2SO_4$ solution = 100.0 mL = 0.1 L

$\text{Moles of} H_2SO_4={0.2}\times{0.1}=0.02moles$

moles of $H^+$ = $2\times 0.02=0.04$

b) Molarity of $NaOH$ solution = 0.5 M

Volume of $NaOH$ solution = 100.0 mL = 0.1 L

$\text{Moles of} NaOH={0.5}\times{0.1}=0.05moles$

moles of $OH^-$ = $1\times 0.05=0.05$

$H^++OH^-\rightarrow H_2O$

According to stoichiometry:

1 mole of $H^+$  combine with 1 mole of $OH^-$

Now 0.04 moles of $H^+$ will combine with 0.04 moles of $OH^-$

Thus heat of neutralization = $0.04\times -57kJ=-2.3kJ$

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