Define mixtures-a H2g - 12 O2g - H2 Ol- - Ho298 K- - 285-9 kJ mol-1b H2g - 12O2g - H2Og- - Ho298 K - -241-8 kJ mol-1The molar enthalpy of vapourisation of water will be-

The correct option is C 44.1 kJ mol^ 1 Given:(a) H_2_( g ) + 12O_2_( g )⟶H_2O_( l ); ΔH = 285.9 kJ/mol H_2O_( l )⟶H_2_( g ) + 12 ...

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Standard XII

Chemistry

Enthalpy

Define mixtur...

Question

Define mixtures.
(a) $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$ ;
$Δ H_{298 K}^{o} = - 285.9 k J m o l^{- 1}$
(b) $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)$ ;
$Δ H_{298 K}^{o} = - 241.8 k J m o l^{- 1}$
The molar enthalpy of vapourisation of water will be:

241.8

k J

m o l^{- 1}

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22.0

k J

m o l^{- 1}

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44.1

k J

m o l^{- 1}

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527.7

k J

m o l^{- 1}

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Solution

The correct option is C $44.1$ $k J$ $m o l^{- 1}$
Given:
(a) ${H_{2}}_{(g)} + \frac{1}{2} {O_{2}}_{(g)} ⟶ {H_{2} O}_{(l)}; Δ H = - 285.9 k J / m o l$
${H_{2} O}_{(l)} ⟶ {H_{2}}_{(g)} + \frac{1}{2} {O_{2}}_{(g)}; Δ H = 285.9 k J / m o l . . . . . (1)$
(b) ${H_{2}}_{(g)} + \frac{1}{2} {O_{2}}_{(g)} ⟶ {H_{2} O}_{(g)}; Δ H = - 241.8 k J / m o l . . . . . (2)$
Adding ${e q}^{n} (1) & (2)$ , we have
${H_{2} O}_{(l)} + {H_{2}}_{(g)} + \frac{1}{2} {O_{2}}_{(g)} ⟶ {H_{2}}_{(g)} + \frac{1}{2} {O_{2}}_{(g)} + {H_{2} O}_{(g)}; Δ H = [285.9 + (- 241.8)] k J / m o l$
${H_{2} O}_{(l)} ⟶ {H_{2} O}_{(g)}; Δ H = 44.1 k J / m o l$
Hence the molar enthalpy of vapourisation of water is $44.1 k J / m o l$ .

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Similar questions

Q. Consider the following reactions.

(a)

H_{(a q)}^{+} + O H_{(a q)}^{-} = H_{2} O_{(l)}, Δ H = - X_{1}

m o l^{- 1}

(b)

H_{2 (g)} + \frac{1}{2} O_{2 (g)} = H_{2} O_{(l)}, Δ H = X_{2}

m o l^{- 1}

(c)

C O_{2 (g)} + H_{2 (g)} = C O_{(g)} + H_{2} O_{(l)} - X_{3}

m o l^{- 1}

(d)

C_{2} H_{2 (g)} + \frac{5}{2} O_{2 (g)} = 2 C O_{2 (g)} + H_{2} O_{(l)} + X_{4}

m o l^{- 1}

Enthalpy of formation of

H_{2} O_{(l)}

is:

Q. Calculate the standard enthalpy of formation of

C H_{3} O H (l)

from the following data.

C H_{3} O H (l) + \frac{3}{2} O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)

;

Δ H^{o} = - 726

m o l^{- 1}

C(graphite)

+ O_{2} (g) \to C O_{2} (g);

Δ_{c} H^{0} = - 393

m o l^{- 1}

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l);

Δ_{t} H^{o} = - 286

m o l^{- 1}

Q. Equal volumes of

C_{2} H_{2}

and

H_{2}

are combusted under identical conditions. The ratio of heat evolved for

C_{2} H_{2}

and

H_{2}

is:

H_{2} (g) + 1 / 2 O_{2} (g) \to H_{2} O (g),

Δ H = - 241.8

kJ/mol

C_{2} H_{2} (g) + 5 / 2 O_{2} (g) \to 2 C O_{2} (g) + H_{2} O (g)

Δ H = - 1300

kJ/mol

Q. The

Δ_{f} H^{o}

for the

C O_{2} (g), C O (g)

and

H_{2} O (l)

are -393.5, -110.5 and -241.8

k J m o l^{- 1}

respectively. The standard enthalpy change (in

k J m o l^{- 1}

) for the given reaction is :

C O_{2} (g) + H_{2} (g) \to C O (g) + H_{2} O (l)

Assuming that water vapour is an ideal gas, the internal energy change (∆U) when 1 mol of water is vapourised at 1 bar pressure and 100 $^{\circ}$ C, (given : molar enthalpy of vapourisation of water at 1 bar and 373 K = 41 kJ mol $^{- 1}$ and R = 8.3 J mol $^{- 1}$ K $^{- 1}$ ) will be

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Define mixtures.(a) H2(g)+12O2(g)→H2O(l); ΔHo298 K=−285.9 kJmol−1(b) H2(g)+12O2(g)→H2O(g); ΔHo298 K=−241.8 kJmol−1The molar enthalpy of vapourisation of water will be:

Define mixtures.
(a) $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$ ;
$Δ H_{298 K}^{o} = - 285.9 k J m o l^{- 1}$
(b) $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)$ ;
$Δ H_{298 K}^{o} = - 241.8 k J m o l^{- 1}$
The molar enthalpy of vapourisation of water will be: