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Calorimetry

At 298 K, the...

Question

At $298 K$ , the enthalpy of fusion of a solid $(X)$ is $2.8 {KJ mol}^{- 1}$ and the enthalpy of vaporisation of the liquid $(X)$ is $98.2 K J {mol}^{- 1}$ . The enthalpy of sublimation of the substance $(X)$ in $K J {mol}^{- 1}$ is . (in nearest integer)

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Solution

$Δ_{sub} H = Δ_{fus} H + Δ_{vap} H$ (These values should be at a given temperature)
$Δ_{sub} H = 98.2 + 2.8 Δ_{sub} H = 101 K J {mol}^{- 1}$

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Similar questions

298 K

(X)

2.8 {KJ mol}^{- 1}

(X)

98.2 K J {mol}^{- 1}

(X)

K J {mol}^{- 1}

K C l

K = 89 k J m o l^{- 1}

C l_{2} = 244 k J m o l^{- 1}

K = 425 k J m o l^{- 1}

C l = - 355 k J m o l^{- 1}

K C l = - 438 k J m o l^{- 1}

30 k J / m o l

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

1 a t m

M g F_{2}

M g (s) \to M g (g) = 146.4 k J m o l^{- 1}

F_{2} (g) \to 2 F (g) = 155.8 k J m o l^{- 1}

M g (g) \to M g^{2 +} (g) = 2186 k J m o l^{- 1}

2 F \to 2 F^{- 1} = 2 \times - 322.6 = - 645.2 k J m o l^{- 1}

M g F_{2} = - 2922.5 k J m o l^{- 1}

k J m o l^{- 1}

L i F

155.3 k J m o l^{- 1}

F_{2} = 75.3 k J

= 520 k J m o l^{- 1}

F (g) = - 333 k J Δ_{f} H_{o v e r a l l} = - 594 k J m o l^{- 1}

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