The heat of neutralisation of $N a O H$ with $H C l$ is $57.3$ kJ and with $H C N$ is $12.1$ kJ. The heat of ionisation of $H C N$ is

+69.4 kJ

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+45.2 kJ

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-45.2 kJ

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-69.4 kJ

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Solution

The correct option is B +45.2 kJ
Heat of ionisation + Heat of neutralisation of $N a O H$ with $H C N$ $=$ Heat of neutralisation of $N a O H$ with $H C l$
Heat of ionisation + $12.1$ KJ $=$ $57.3$ KJ
$∴$ Heat of ionisation $= 57.3 - 12.1$ KJ $= 45.2$ KJ

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Q. If

Δ H_{i o n i z} (H C N) = 45.2 k J m o l^{- 1}

Δ H (C H_{3} C O O H) = 2.1 k J / m o l

then:

Given $Δ H_{i o n i n a z}^{0}$ (HCN) = 45.2 kJ mol $^{- 1}$ and $Δ H_{i o n i n a z}^{0}$ (CH $_{3}$ COOH) = 2.1 kJ mol $^{- 1}$ . Which one of the following is true?

Q. Heat of neutralization of NaOH and HCl is -57.46 kJ/equivalent. The heat of ionisation of water in kJ/mol is:

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The heat of neutralisation of NaOH with HCl is 57.3 kJ and with HCN is 12.1 kJ. The heat of ionisation of HCN is

The correct option is B +45.2 kJHeat of ionisation + Heat of neutralisation of NaOH with HCN = Heat of neutralisation of NaOH with HClHeat of ionisation + 12.1 KJ = 57.3 KJ∴ Heat of ionisation =57.3−12.1 KJ =45.2 KJ

The heat of neutralisation of $N a O H$ with $H C l$ is $57.3$ kJ and with $H C N$ is $12.1$ kJ. The heat of ionisation of $H C N$ is

The correct option is B +45.2 kJ
Heat of ionisation + Heat of neutralisation of $N a O H$ with $H C N$ $=$ Heat of neutralisation of $N a O H$ with $H C l$
Heat of ionisation + $12.1$ KJ $=$ $57.3$ KJ
$∴$ Heat of ionisation $= 57.3 - 12.1$ KJ $= 45.2$ KJ