Question

100ml of 1Mof HCl is mixed with 200ml of 0.5Mof koh enthalpy change(kj) in this process will be​

Answer 1

Answer:

the ans is -5.72kj

Explanation:

The net equation representing the neutralization of a strong acid and a strong base,

H^{+} (aq) + OH^{-}(aq) -- > H_{2}O (l)H

+

(aq)+OH

−

(aq)−−>H

2

O(l)

Standard heat of neutralization for the titration of a strong acid and a strong base is -57.2 kJ/mol

Calculating the moles of acid :

1 \frac{mol}{L} HCl * 100 mL * \frac{1 L}{1000 mL} = 0.1 mol1

L

mol

HCl∗100mL∗

1000mL

1L

=0.1mol

Calculating the moles of base:

0.5\frac{mol}{L} * 200 mL * \frac{1 L}{1000 mL} = 0.1 mol0.5

L

mol

∗200mL∗

1000mL

1L

=0.1mol

The enthalpy change in the above neutralization reaction will be,

0.1 mol * \frac{-57.1 kJ}{1 mol} = -5.71 kJ0.1mol∗

1mol

−57.1kJ

=−5.71kJ

Therefore the correct answer is c. -5.72 kJ