Calculate the DHof for the following reaction under standard state conditions:
2 Ag2S(s) + 2 H2O(l) ---> 4 Ag(s) + 2 H2S(g) + O2(g)
Given the following information:
Ag2S(s) ÄHof = -32.6 kJ/mol
H2O(l) Ä Hof = -285.83 kJ/mol
H2S(g) ÄHof = -20.6 kJ/mol
Question options:
309.8 kJ
595.7 kJ
-595.7 kJ
297.8 kJ
-297.8 kJ
595 kJ
Calculate the DHof for the following reaction under standard state conditions:
2 Ag2S(s) + 2 H2O(l) ---> 4 Ag(s) + 2 H2S(g) + O2(g)
Given the following information:
Ag2S(s) ÄHof = -32.6 kJ/mol
H2O(l) Ä Hof = -285.83 kJ/mol
H2S(g) ÄHof = -20.6 kJ/mol
To calculate the ΔHof (standard enthalpy change of formation) for the given reaction, we need to subtract the sum of the enthalpies of formation of the reactants from the sum of the enthalpies of formation of the products.
Given information:
Ag2S(s) ΔHof = -32.6 kJ/mol
H2O(l) ΔHof = -285.83 kJ/mol
H2S(g) ΔHof = -20.6 kJ/mol
Reactants:
2 Ag2S(s) + 2 H2O(l)
Products:
4 Ag(s) + 2 H2S(g) + O2(g)
Now, let's calculate the ΔHof for the given reaction:
Reactants:
2 Ag2S(s) = 2 × (-32.6 kJ/mol) = -65.2 kJ/mol
2 H2O(l) = 2 × (-285.83 kJ/mol) = -571.66 kJ/mol
Products:
4 Ag(s) = 4 × 0 kJ/mol = 0 kJ/mol
2 H2S(g) = 2 × (-20.6 kJ/mol) = -41.2 kJ/mol
O2(g) = 0 kJ/mol (since it is an element in its standard state)
ΔHof = (sum of products' enthalpies of formation) - (sum of reactants' enthalpies of formation)
ΔHof = (0 kJ/mol + (-41.2 kJ/mol) + 0 kJ/mol) - (-65.2 kJ/mol + (-571.66 kJ/mol))
ΔHof = -41.2 kJ/mol + 65.2 kJ/mol + 571.66 kJ/mol
ΔHof = 595.66 kJ/mol
Therefore, the ΔHof for the given reaction under standard state conditions is 595.7 kJ/mol.
The correct answer is:
595.7 kJ
To calculate the change in enthalpy (ΔH°f) for the given reaction using the standard enthalpy of formation (ΔH°f) values, you need to follow these steps:
1. Write the balanced chemical equation for the reaction:
2 Ag2S(s) + 2 H2O(l) → 4 Ag(s) + 2 H2S(g) + O2(g)
2. Determine the ΔH°f of the products and reactants. The ΔH°f values given are:
ΔH°f(Ag2S(s)) = -32.6 kJ/mol
ΔH°f(H2O(l)) = -285.83 kJ/mol
ΔH°f(H2S(g)) = -20.6 kJ/mol
3. Calculate the ΔH°f for the reaction by summing the ΔH°f of the products and subtracting the sum of the ΔH°f of the reactants. Since the stoichiometric coefficients in the balanced equation are 2, we double the ΔH°f values of the reactants:
ΔH°f(reaction) = (4 * ΔH°(s))) + (2 * ΔH°f(H2S(g))) + ΔH°f(O2(g)) - (2 * ΔH°f(Ag2S(s))) - (2 * ΔH°f(H2O(l)))
Substituting the given values:
ΔH°f(reaction) = (4 * 0) + (2 * (-20.6)) + 0 - (2 * (-32.6)) - (2 * (-285.83))
ΔH°f(reaction) = (-41.2) + 65.2 + 32.6 + 571.66
ΔH°f(reaction) = 627.66 kJ/mol
Therefore, the ΔH°f for the given reaction under standard state conditions is 627.66 kJ/mol. However, none of the options provided match this value.