Question

The enthalpy change, ΔH, for a reaction at constant pressure is defined as: ΔH = ΔE + PΔV. For which of the following reactions will ΔH be approximately equal to ΔE? Select all that apply. Group of answer choices 2 NO2(g) -> N2(g) + 2 O2(g)

Answer 1

The given question is incomplete. The complete question is:

The enthalpy change, ΔH, for a reaction at constant pressure is defined as: ΔH = ΔE + PΔV. For which of the following reactions will ΔH be approximately equal to ΔE? Select all that apply. Group of answer choices

$2NO_2(g)\rightarrow N_2(g)+2O_2(g)$

$Ca(OH)_2(aq)+H_2SO_4(aq)\rightarrow 2H_2O(l)+caSO_4(s)$

$C(s)+O_2(g)\rightarrow CO_2(g)$

None of the above

Answer:  

$C(s)+O_2(g)\rightarrow CO_2(g)$

Explanation:

Relation of  with  is given by the formula:

$\Delta H=\Delta E+{\Delta n_g}RT$     as

Where,

$\Delta H$ = enthalpy change

$\Delta E$= internal energy change

R = Gas constant

T = temperature

$\Delta n_g$= change in number of moles of gas particles = $n_{products}-n_{reactants}$

1. For $2NO_2(g)\rightarrow N_2(g)+2O_2(g)$

$\Delta n_g=n_{products}-n_{reactants}=(3-2)=1$

2. For $Ca(OH)_2(aq0+H_2SO_4(aq)\rightarrow 2H_2O(l)+CaSO_4(s)$

$\Delta n_g=n_{products}-n_{reactants}=(0-0)=0$

3.  $C(s)+O_2(g)\rightarrow CO_2(g)$

$\Delta n_g=n_{products}-n_{reactants}=(0-0)=0$

Thus for reactions 2 and 3,  ΔH be approximately equal to ΔE