When MgO reacts with H2O at 25° C and 1 atm, the volume change is −4.6mL⋅mol−1. MgO(s)+H2O(l)⟶Mg(OH)2(s)What is the value of ΔH−ΔE for this reaction?

− 4.7 × 10 − 1 J ⋅ mol − 1

− 4.7 × 10 − 1 J ⋅ mol − 1

− 4.7 × 10 2 J ⋅ mol − 1

4.7 × 10 − 1 J ⋅ mol − 1

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When MgO reacts with $H_{2} O$ at 25° C and 1 atm, the volume change is $- 4.6 mL \cdot {mol}^{- 1}$ .
$MgO (s) + H_{2} O (l) ⟶ Mg (OH)_{2} (s)$
What is the value of $Δ H - Δ E$ for this reaction?

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$- 4.7 \times 10^{- 1} J \cdot {mol}^{- 1}$

$- 4.7 \times 10^{2} J \cdot {mol}^{- 1}$

$4.7 \times 10^{2} J \cdot {mol}^{- 1}$

$4.7 \times 10^{- 1} J \cdot {mol}^{- 1}$

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The difference between the change in enthalpy and change in internal energy is given as,

$\begin{array}{l} Δ H - Δ E = Δ (P V) \\ = - 4.6 \times 10^{- 3} \times 1 \times 101.3 \\ = - 4.7 \times 10^{- 1} \end{array}$

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