Respuesta :
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
[tex]2NO_2(g)\rightarrow N_2(g)+2O_2(g)[/tex]
[tex]Ca(OH)_2(aq)+H_2SO_4(aq)\rightarrow 2H_2O(l)+caSO_4(s)[/tex]
[tex]C(s)+O_2(g)\rightarrow CO_2(g)[/tex]
None of the above
Answer:
[tex]C(s)+O_2(g)\rightarrow CO_2(g)[/tex]
Explanation:
Relation of with is given by the formula:
[tex]\Delta H=\Delta E+{\Delta n_g}RT[/tex] as
Where,
[tex]\Delta H[/tex] = enthalpy change
[tex]\Delta E[/tex]= internal energy change
R = Gas constant
T = temperature
[tex]\Delta n_g[/tex]= change in number of moles of gas particles = [tex]n_{products}-n_{reactants}[/tex]
1. For [tex]2NO_2(g)\rightarrow N_2(g)+2O_2(g)[/tex]
[tex]\Delta n_g=n_{products}-n_{reactants}=(3-2)=1[/tex]
2. For [tex]Ca(OH)_2(aq0+H_2SO_4(aq)\rightarrow 2H_2O(l)+CaSO_4(s)[/tex]
[tex]\Delta n_g=n_{products}-n_{reactants}=(0-0)=0[/tex]
3. [tex]C(s)+O_2(g)\rightarrow CO_2(g)[/tex]
[tex]\Delta n_g=n_{products}-n_{reactants}=(0-0)=0[/tex]
Thus for reactions 2 and 3, ΔH be approximately equal to ΔE
The options in which ΔH be approximately equal to ΔE are:
(ii) and (iii)
Enthalpy change:
The enthalpy change, ΔH, for a reaction at constant pressure is defined as:
[tex]\triangle H = \triangle E + \triangle n_g RT\\\\\triangle n_g=n_{\text{products}} -n_{\text{reactants}}[/tex]
Now, for each given choice we will calculate the change in number of moles of gas particles. Since, ΔH, for a reaction at constant pressure is defined as: ΔH = ΔE + PΔV
(i) [tex]2NO_2(g)---->N_2(g)+2O_2(g)[/tex]
[tex]\triangle n_g=n_{\text{products}} -n_{\text{reactants}}\\\\\triangle n_g=3-2=1[/tex]
(ii)[tex]Ca(OH)_2(aq)+H_2SO_4(aq)--->2H_2O(l)+CaSO_4(s)[/tex]
[tex]\triangle n_g=n_{\text{products}} -n_{\text{reactants}}\\\\\triangle n_g=0-0=0[/tex]
(iii) [tex]C(s)+O_2(g)---->CO_2(g)[/tex]
[tex]\triangle n_g=n_{\text{products}} -n_{\text{reactants}}\\\\\triangle n_g=0-0=0[/tex]
Thus for reactions (ii) and (iii) , ΔH be approximately equal to ΔE.
Therefore, correct options are (ii) and (iii).
Find more information about Enthalpy change here: brainly.com/question/14047927
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
(i) [tex]2NO_2(g)---->N_2(g)+2O_2(g)[/tex]
(ii) [tex]Ca(OH)_2(aq)+H_2SO_4(aq)--->2H_2O(l)+CaSO_4(s)[/tex]
(iii) [tex]C(s)+O_2(g)---->CO_2(g)[/tex]
(iv) None of the above
