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Heat of Reaction

Find the valu...

Question

Find the value of $Δ E$ when 2.0 mole of liquid water vaporises at $100^{o} C$ is:
Given: The heat of vaporisation $(Δ H v a p .)$ of water at $100^{o} C$ is $40.66 k J m o l^{- 1}$

Δ E = 75.11 k J

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Δ E = - 75.11 k J

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Δ E = 87.5 k J

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none of the above

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Solution

The correct option is A $Δ E = 75.11 k J$
$q_{p} = q_{v} + Δ n g R T$
$40.66 \times 2 = q_{v} + 2 \times 8.314 \times 373$
$\Rightarrow q - v = (81.32 - 6.202) k J$
$q_{v} = 75.118 k J$

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Similar questions

Δ E

2.0

100^{o} C

Δ H

100^{o} C

40.66 k J {m o l}^{- 1}

Δ U

2.0

100^{o} C

(Δ H_{v a p})

100^{o} C

40.66 K J {m o l}^{- 1}

Assertion(A): The increase in internal energy ( $Δ E$ ) for the vaporisation of one mole of water at 1atm and 373K is zero.

Reason(R): For all isothermal process, $Δ E = 0$

100^{\circ} C

40.63 k J m o l^{- 1}

Δ E

△ U

100^{\circ} ?

△ H

100^{\circ} C

40.66

k J m o l^{- 1}

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