Answer: d) -705.55 kJ
Explanation:
Heat of reaction is the change of enthalpy during a chemical reaction with all substances in their standard states.
Reversing the reaction, changes the sign of 
On multiplying the reaction by 
, enthalpy gets half:
Thus the enthalpy change for the given reaction is -705.55kJ
The balanced equation for the decomposition of solid lead iv oxide is as follows: 2PbO2 = 2PbO + O2.
Lead IV oxide decompose to give lead ll oxide and oxygen. Lead iv oxide is thermally unstable and it usually decomposes into oxygen and lead ll oxide when heated. Lead ll oxide is more stable than lead lV oxide.
D. The number increases and then decreases for noble gases
Answer:
D. The equipment needed to accommodate the high temperature and pressure will be expensive to produce.
Explanation:
Hello!
In this case, for the considered reaction, it is clear it is an exothermic reaction because it produces energy; and therefore, the higher the temperature the more reactants are yielded as the reverse reaction is favored. Moreover, since the effect of pressure is verified as favoring the side with fewer moles; in this case the products side (2 moles of ammonia).
In such a way, the high pressure favors the formation of ammonia whereas the high temperature the formation of hydrogen and nitrogen and therefore, option A is ruled out. Since the high pressure shifts the reaction rightwards and the high temperature leftwards, we would not be able to know whether the reaction has ended or not because it will be a "go and come back" process, that is why B is also discarded. Now, since hydrogen and nitrogen would be the "wastes", we discard C because they are not toxic. That is why the most accurate answer would be D. because it is actually true that such equipment is quite expensive.
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Answer:
-138.9 kJ/mol
Explanation:
Step 1: Convert 235.8°C to the Kelvin scale
We will use the following expression.
K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K
Step 2: Calculate the standard enthalpy of reaction (ΔH°)
We will use the following expression.
ΔG° = ΔH° - T.ΔS°
ΔH° = ΔG° / T.ΔS°
ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K
ΔH° = -3.583 kJ (for 1 mole of balanced reaction)
Step 3: Convert -9.9°C to the Kelvin scale
K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K
Step 4: Calculate ΔG° at 263.3 K
ΔG° = ΔH° - T.ΔS°
ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K
ΔG° = -138.9 kJ/mol
