Question

Observe the following reaction, $ HCl + \,NaOH\, \to \,NaCl\, + \,{H_2}O\,\,;\,\,\Delta H = \, - 57.1\,kJ\,mo{l^{ - 1}}\, $
 $ \Delta H $ is the heat of neutralisation of $ HCl $ and $ NaOH $ solution. If true enter $ 1 $ else $ 0 $.

Answer 1

Hint: $ \Delta H $ is the amount of energy that will release when neutralisation reaction happens. Neutralisation reaction is the reaction when an acid and a base reacts and give a salt at the end. The amount of heat released is called its heat of neutralisation. It will depend upon the nature of acid and bases that we use. There is a certain amount of heat of neutralisation when a strong acid base neutralises and different for weak acid bases.

Complete step by step solution:
For the above reaction where a strong acid like hydrochloric acid $ HCl $ will reacts with a strong base like sodium hydroxide $ NaOH $ the heat which will release is of amount $ - 57.1\,kJ\,mo{l^{ - 1}}\, $ . Now, there are some changes that happen when one of the species is weak, like if the neutralisation happens between a weak acid and strong base or weak base and strong acid. During these types of neutralisation reaction the heat of neutralisation is $ 13.4\,kJ\,mo{l^{ - 1}}\, $
The reason behind this difference of energy of neutralisation is the difference in their ionisation. See, when a strong acid and base reacts they will ionise completely and thus the number of ions means hydrogen ion and hydroxide ion will be the same after the reaction. But in case of weak acid base reaction or reaction where one species is weak, strong species get ionised properly while the other one will not ionise. Thus they don’t maintain an enthalpy of $ - 57.1\,kJ\,mo{l^{ - 1}}\, $ . So, you have to enter $ 1 $ for true.

Note :
If you have a reaction between say, ammonium hydroxide $ N{H_4}OH $ and $ HCl $ you have to see what type of electrolyte you have. As we know that $ HCl $ is strong acid while $ N{H_4}OH $ is a weak base, so during the neutralization reaction, one will get fully ionize and other remains partially ionize and hence they will give $ 13.4\,kJ\,mo{l^{ - 1}}\, $ heat of neutralisation.