Given the following thermochemical data:
H2O(l)+2AgNO3(aq) ¡æ 2HNO3(aq)+Ag2O(s) ¥ÄH = 44.8 kJ
2Ag(s)+¨öO2(g) ¡æ Ag2O(s) ¥ÄH = -31.0 kJ
¨öO2(g)+H2(g) ¡æ H2O(l) ¥ÄH = -285.8 kJ
Use Hess¡¯s Law to determine ¥ÄH for the reaction:
¨öH2(g)+AgNO3(aq) ¡æ Ag(s)+HNO3(aq)
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To determine ΔH for the reaction:
½H2(g) + AgNO3(aq) → Ag(s) + HNO3(aq)
We can use Hess's Law and the given thermochemical equations.
Step 1:
Reverse the first equation:
2HNO3(aq) + Ag2O(s) → H2O(l) + 2AgNO3(aq)
Step 2:
Multiply the second equation by 2:
4Ag(s) + O2(g) → 2Ag2O(s)
Step 3:
Multiply the third equation by ½:
½O2(g) + ½H2(g) → ½H2O(l)
Step 4:
Add the modified equations together:
2HNO3(aq) + Ag2O(s) → H2O(l) + 2AgNO3(aq) ΔH = 44.8 kJ (eqn 1)
4Ag(s) + O2(g) → 2Ag2O(s) ΔH = -31.0 kJ (eqn 2)
½O2(g) + ½H2(g) → ½H2O(l) ΔH = -285.8 kJ (eqn 3)
When we add these equations, the Ag2O(s) on the left side of equation 1 will cancel the Ag2O(s) on the right side of equation 2.
2HNO3(aq) + 4Ag(s) + ½O2(g) + ½H2(g) → H2O(l) + 2AgNO3(aq) + 2Ag2O(s) + ½H2O(l)
Simplifying the equation:
2HNO3(aq) + 4Ag(s) + ½O2(g) + ½H2(g) → 2H2O(l) + 2AgNO3(aq) + 2Ag2O(s)
Step 5:
Calculate the net ΔH:
ΔH = (44.8 kJ) + (-31.0 kJ) + (-285.8 kJ)
= -271.6 kJ
Therefore, ΔH for the reaction ½H2(g) + AgNO3(aq) → Ag(s) + HNO3(aq) is -271.6 kJ.
To determine the change in enthalpy (ΔH) for the given reaction, which involves the formation of Ag(s) and HNO3(aq), we can use Hess's Law. Hess's Law states that the overall enthalpy change of a reaction is independent of the pathway taken and depends only on the initial and final conditions.
We can break down the given reaction into a series of steps for which enthalpy changes are known:
1. Multiply the first equation by 2 to obtain:
2H2O(l) + 4AgNO3(aq) → 4HNO3(aq) + 2Ag2O(s) ΔH = 89.6 kJ
2. Multiply the second equation by 2 to obtain:
4Ag(s) + ½O2(g) → 2Ag2O(s) ΔH = -62.0 kJ
3. Reverse the third equation:
H2O(l) → ½O2(g) + H2(g) ΔH = +285.8 kJ
Now, to obtain the target equation, we will add the above equations:
2H2O(l) + 4AgNO3(aq) → 4HNO3(aq) + 2Ag2O(s) ΔH = 89.6 kJ
4Ag(s) + ½O2(g) → 2Ag2O(s) ΔH = -62.0 kJ
H2O(l) → ½O2(g) + H2(g) ΔH = +285.8 kJ
By canceling out species that appear on both sides, we can eliminate H2O(l) and Ag2O(s):
4AgNO3(aq) + 4Ag(s) → 4HNO3(aq) ΔH = 89.6 kJ - 62.0 kJ + 285.8 kJ
4AgNO3(aq) + 4Ag(s) → 4HNO3(aq) ΔH = +313.4 kJ
Now, to determine ΔH for the target equation:
¨öH2(g) + AgNO3(aq) → Ag(s) + HNO3(aq)
We can see that the coefficients are halved compared to the previous equation, so we need to halve the enthalpy change as well:
¨öH2(g) + AgNO3(aq) → Ag(s) + HNO3(aq) ΔH = ½(313.4 kJ)
¨öH2(g) + AgNO3(aq) → Ag(s) + HNO3(aq) ΔH = 156.7 kJ
Therefore, the change in enthalpy (ΔH) for the reaction ¨öH2(g) + AgNO3(aq) → Ag(s) + HNO3(aq) is 156.7 kJ.