The enthalpy change (Δ H) for the neutralization of M HCl by caustic potash in dilute solution at 298 K is:

Question

The enthalpy change (Δ H) for the neutralization of M HCl by caustic potash in dilute solution at 298 K is: (A) 68 kJ (B) 65 kJ (C) 57.3 kJ (D) 50 kJ

Tardigrade Admin · Accepted Answer

underset beginsmallmatrix (strong acid endsmallmatrix mathopHCl + underset beginsmallmatrix (strong base) endsmallmatrix mathopKOH KCl+H2O In this reaction, HCl is the .strong acid and KOH is the base and it has been found that the heat of neutralization of a strong acid with a strong base is always constant i. e., 57.3 kJ.

Q. The enthalpy change $(Δ H)$ for the neutralization of M $H Cl$ by caustic potash in dilute solution at 298 K is:

68 kJ

65 kJ

57.3 kJ

50 kJ

$(s t ro n g a c i d H Cl + (s t ro n g ba se) K O H K Cl + H_{2} O$ In this reaction, $H Cl$ is the .strong acid and $K O H$ is the base and it has been found that the heat of neutralization of a strong acid with a strong base is always constant i. e., 57.3 kJ.

Q. The enthalpy change (ΔH) for the neutralization of M HCl by caustic potash in dilute solution at 298 K is: