The enthalpy of vaporisation of liquid water using the data `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l),DeltaH=-285.77kJ//mol` `H_(2)(g)+(1)/(2)O_(2)(g)toH_(

Question

The enthalpy of vaporisation of liquid water using the data `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l),DeltaH=-285.77kJ//mol` `H_(2)(g)+(1)/(2)O_(2)(g)toH_(

The enthalpy of vaporisation of liquid water using the data
`H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l),DeltaH=-285.77kJ//mol`
`H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(g),DeltaH=-241.84kJ//mol`
is
A. `+43.93kJ//mol`
B. `-43.93kJ//mol`
C. `527.61kJ//mol`
D. `-527.61kJ//mol`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

Related Questions

Standard enthalpy of vapourisation `DeltaH^(@)` forwater is `40.66 KJ mol^(-1)`.The internal energy of vapourisation of water for its 2 mol will be :-

2 Answers 4 Views

Consider the following process :- `Ato2B,DeltaH=+150 KJ` `3Bto2C+D,DeltaH=+350 KJ` `G+Ato2D,DeltaH=+350 KJ` For `B+DtoG+2C,DeltaH` will be

2 Answers 4 Views

Given `C_(s)+O_(2(g))toCO_(2(g)),DeltaH=-94.2K cal` `H_(2(g))+½O_(2(g))toH_(2)O_((g)),Delta=-68.3K cal` `CH_(4(g))+2O_(2(g))+2H=-210.8K cal` what will

2 Answers 4 Views

Calculate the change in entropy for the process : \( H _{2} O _{\left(\text {liq, } 20^{\circ} C , 1 atm \right)} \) to \( H _{2} O _{\left.\text {(gas, } 250^{\circ} C , 1 atm \right)} \) You are given, heat capacity of water \( ( \) liq \( )=75.6 JK ^{-1} \cdot mol ^{-1} \) Heat capacity of water \( ( \) vap \( )=36.2 JK ^{-1} mol ^{-1} \) Heat of vaporization of \( H _{2} O \) at \( 100^{\circ} C \) and \( 1 atm 40.85 JK mol ^{-1} \)

2 Answers 4 Views

Molar enthalpy change for vaporisation of 1 mole of water at 1 bar and `100^(@)C` is 41 kJ `mol^(-1)` (If water vapour is assumed to be perfect gas).

2 Answers 5 Views

Given, `2Fe+(3)/(2)O_(2)toFe_(2)O_(3),DeltaH=-193.4kJ,Mg+(1)/(2)O_(2)to MgO,DeltaH=-140.2kJ` What is the `DeltaH` for the following reaction? `3Mg+Fe_

2 Answers 4 Views

Calculate the enthalpy of vaporisation of `I_(2)` if the sublimation energy and enthalpy of fusion of `I_(2)` is 57.3 kJ `mol^(-1)` and 15.5 kJ `mol^(

2 Answers 4 Views

For the reaction `AtoB,DeltaH=+23kJ//mol and BtoC,DeltaH=-18kJ//mol`, the decreasing order of enthalpy of A,B,C follows the order

2 Answers 4 Views

The `DeltaH` for vaporization of a liquid is 20kJ/mol. Assuming ideal behaviour, the change in internal energy for the vaporization of 1 mol of the li

2 Answers 4 Views

given `NO(g)+O_(3)(g)toNO_(2)(g)+O_(2)(g) " "DeltaH=198.9kJ//mol` `O_(3)(g)to3//2O_(2)(g) " " DeltaH=-142.3 kj//mol` `O_(2)(g)to2O(g) " " DeltaH=+495.

2 Answers 6 Views

Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - A
(I) `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l),DeltaH=-285.77kJ//mol`
(II) `H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(g),DeltaH=-241.84kJ//mol`
`H_(2)O(l)toH_(2)O(g),dDeltaH=?`
Subtracting the Eqs. (I) from (II), we get
`H_(2)O(l)toH_(2)O(g)`
`DeltaH=-241.84-(-285.77)`
`DeltaH=+43.93kJ//mol`

4 views Answered 2023-02-05 15:29