The standard enthalpy change for the neutralisation of hydrocyanic acid and NaOH is Δ H° at 298=-2460 cal. Given that the standard enthalpy change for the neutralization of HCl and NaOH is Δ H=-13.71 kcal. What is the heat of ionization of HCN ?

Question

The standard enthalpy change for the neutralisation of hydrocyanic acid and NaOH is Δ H° at 298=-2460 cal. Given that the standard enthalpy change for the neutralization of HCl and NaOH is Δ H=-13.71 kcal. What is the heat of ionization of HCN ? (A) +15820 cal (B) -11250 cal (C) +11250 cal (D) -15820 cal

Tardigrade Admin · Accepted Answer

Heat of neutralization of HCN = Total heat - heat of neutralization of strong base =-2.460-(-13.71)=11.25 kcal or 11250 cal

Q. The standard enthalpy change for the neutralisation of hydrocyanic acid and $N a O H$ is $Δ H^{\circ}$ at $298 = - 2460 c a l$ . Given that the standard enthalpy change for the neutralization of $H Cl$ and $N a O H$ is $Δ H = - 13.71 k c a l$ . What is the heat of ionization of $H CN ?$

$+ 15820 c a l$

$- 11250 c a l$

$+ 11250 c a l$

$- 15820 c a l$

Heat of neutralization of $H CN$

= Total heat - heat of neutralization of strong base

$= - 2.460 - (- 13.71) = 11.25 k c a l$ or $11250 c a l$

Q. The standard enthalpy change for the neutralisation of hydrocyanic acid and NaOH is ΔH∘ at 298=−2460cal. Given that the standard enthalpy change for the neutralization of HCl and NaOH is ΔH=−13.71kcal. What is the heat of ionization of HCN?

+15820cal

−11250cal

+11250cal

−15820cal