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The heat of neutralisation of oxalic aci...

The heat of neutralisation of oxalic acid is `-25.4 kcal mol^(-1)` using strong base, `NaOH`. Hence, the enthalpy change of the process is
`H_(2)C_(2)O_(4) hArr 2H^(o+) +C_(2)O_(4)^(2-)` is

A

`a//b`

B

`sqrt(a)//2b`

C

`a^2//2b`

D

`a^2//4b`

Text Solution

Verified by Experts

The correct Answer is:
A
`y=ax-bx^(2))`
Range is solution of y=0 `rArr ax-bx^(2)=0 rArr x=a//b rArr" Range "=a//b`
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Heat of neutralization between HCl and NaOH is -13.7 kcal "equiv"^(–1) . Heat of neutralization of H_2C_2O_4 (oxalic acid) with NaOH is -26 kcal mol^(-1) . Hence, heat of dissociation of H_(2)C_(2)O_(4) as H_2C_2O_(4) rarr 2H^(+) + C_(2)O_(4)^(2-) , is :

Heat of neutralisation of oxalic acid is -106.7 KJ mol^(-1) using NaOH hence Delta H of : H_(2)C_(2)O_(4)rarr C_(2)O_(4)^(2-)+2H^(+) is :-

SrC_(2)O_(4)darr+2HCl to SrCl_(2)+H_(2)C_(2)O_(4)

SrC_(2)O_(4)darr+2HCl to SrCl_(2)+H_(2)C_(2)O_(4)

Review the equilibrium and choose the correct statement HCIO_(4) + H_(2)O hArr H_(2)O^(+) CIO_(4)^(-)

BaC_(2)O_(4)darr+2AcOH to Ba(AcO)_(2)+H_(2)C_(2)O_(4)

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It is known that entropy of neutralisation of a strong acid and strong base is -13.68kcal/mol.If enthalpy of neutralisation of N_2H_4 (hydrazine) with strong acid is equal to -11.68 kcal/mol, then enthalpy of the reaction N_2H_4+H_2OtoN_2H_5^+(aq)+OH^(-)(aq) in Kcal/mol is :

Heat of formation of H_(2)O is -188kJ/mol and H_(2)O_(2) is -286 kJ/mol. The enthalpy change for the reaction, 2H_(2)O_(2) to 2H_(2)O+O_(2)

Calculate the heat of formation of methane in kcalmol^(-1) using the following thermo chemical reactions C(s)+O_(2)toCO_(2)(g) , DeltaH=-94.2 kcal mol^(-1) H_(2)(g)+1/2O_(2)(g)toH_(2)O(l) , DeltaH=-68.3 kcal mol^(-1) CH_(4)(g)+2O_(2)(g)toCO_(2)(g)+2H_(2)O(l) , DeltaH=-210.8 kcal mol^(-1)

VMC MODULES ENGLISH-Motion in Two Dimensions-Level -1
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