Calculate the change in enthalpy for the following process at 1 atm H2Ol-50oC-H2Og-150oCgiven that -Hv at 100oC is 40-7kJmol-1 CPH2O-l-75-0Jmol-1K-1 CPH2O-g-33-3Jmol-1K-1

The correct option is A 46.1KJ H_2O(l,50^∘C)→ H_2O(l,100^∘C) H_1=C_p(100 50)=75× 50=3.75 kJ mol^ 1 H_2O(l,100^∘C)→ H_2O(g,100^∘C) H_2 ...

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Byju's Answer

Standard XII

Chemistry

Bond Order and Enthalpy

Calculate the...

Question

Calculate the change in enthalpy for the following process at $1$ atm
$H_{2} O (l, 50^{o} C) \to H_{2} O (g, 150^{o} C)$
given that $Δ H_{v}$ at $100^{o} C$ is $40.7 k J {m o l}^{- 1}$
$C_{P} (H_{2} O, l) = 75.0 J {m o l}^{- 1} K^{- 1}$
$C_{P} (H_{2} O, g) = 33.3 J {m o l}^{- 1} K^{- 1}$

64.8 K J

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52.5 K J

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48.6 K J

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46.1 K J

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Solution

The correct option is A $46.1 K J$
$H_{2} O (l, 50^{\circ} C) \to H_{2} O (l, 100^{\circ} C)$
$△ H_{1} = C_{p} (100 - 50) = 75 \times 50 = 3.75 k J m o l^{- 1}$
$H_{2} O (l, 100^{\circ} C) \to H_{2} O (g, 100^{\circ} C)$
$△ H_{2} = △ H_{v} = 40.7 k J m o l^{- 1}$
$H_{2} O (g, 100^{\circ} C) \to H_{2} O (g, 150^{\circ} C)$
$△ H_{3} = C_{p} (150 - 100) = 33.3 \times 50 = 1.665 k J m o l^{- 1}$
Let no. of moles $= 1$
$∴ △ H = △ H_{1} + △ H_{2} + △ H_{3}$
$= 40.7 + 3.75 + 1.665 = 46.1 k J$

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Calculate the change in enthalpy for the following process at 1 atmH2O(l,50oC)→H2O(g,150oC)given that ΔHv at 100oC is 40.7kJmol−1CP(H2O,l)=75.0Jmol−1K−1CP(H2O,g)=33.3Jmol−1K−1

The correct option is A 46.1KJH2O(l,50∘C)→H2O(l,100∘C)△H1=Cp(100−50)=75×50=3.75kJmol−1H2O(l,100∘C)→H2O(g,100∘C)△H2=△Hv=40.7kJmol−1H2O(g,100∘C)→H2O(g,150∘C)△H3=Cp(150−100)=33.3×50=1.665kJmol−1Let no. of moles=1∴△H=△H1+△H2+△H3=40.7+3.75+1.665=46.1kJ

Calculate the change in enthalpy for the following process at $1$ atm
$H_{2} O (l, 50^{o} C) \to H_{2} O (g, 150^{o} C)$
given that $Δ H_{v}$ at $100^{o} C$ is $40.7 k J {m o l}^{- 1}$
$C_{P} (H_{2} O, l) = 75.0 J {m o l}^{- 1} K^{- 1}$
$C_{P} (H_{2} O, g) = 33.3 J {m o l}^{- 1} K^{- 1}$