A piece of wood near a fire is at 23°C. It gains 1,160 joules of heat from the fire and reaches a temperature of 42°C. The speci
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<u>Answer:</u> The volume of hydrogen gas collected over water is 2.13 L
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
Given mass of zinc = 5.566 g
Molar mass of zinc = 65.4 g/mol
Putting values in above equation, we get:
For the given chemical reaction:
As, HCl is present in excess. So, it is considered as an excess reagent.
Zinc is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of zinc produces 1 mole of hydrogen gas.
So, 0.0851 moles of zinc will produce = of hydrogen gas
To calculate the volume of hydrogen gas, we use ideal gas equation, which is:
PV = nRT
where,
P = Pressure of hydrogen gas = Total atmospheric pressure - vapor pressure of water = (752 - 18.65) mmHg = 733.35 mmHg
V = Volume of the hydrogen gas
n = number of moles of gas = 0.0851 moles
R = Gas constant =
T = Temperature of hydrogen gas =
Putting values in above equation, we get:
Hence, the volume of hydrogen gas collected over water is 2.13 L
The question is incomplete, complete question is :
Consider the decomposition of red mercury(II) oxide under standard state conditions.
Given :
Enthalpy change of the reaction = ΔH = -90.83 kJ/mol
(a) Is the decomposition spontaneous under standard state conditions?
(b) Above what temperature does the reaction become spontaneous?
Answer:
a) The decomposition is spontaneous under standard state conditions.
b)The reaction will spontaneous above -419.69 Kelvins.
Explanation:
Given :
Entropy change of the reaction ; ΔS
Enthalpy change of the reaction = ΔH = -90.83 kJ/mol = -90830 J/mol K
1 kJ = 1000 J
At standard condition the value of temperature = T = 298 K
ΔG = ΔH - TΔS
ΔG = -90830 J/mol K - 298 K × 216.42 J/mol K = -155,323.16 J/mol
ΔG < 0 ( spontaneous)
The decomposition is spontaneous under standard state conditions.
b) Above what temperature does the reaction become spontaneous
Let the ΔG = 0
Enthalpy change of the reaction = ΔH = -90.83 kJ/mol = -90830 J/mol K
ΔG = ΔH - TΔS
0 = -90830 J/mol K - T × 216.42 J/mol K
T = -419.69 K
The reaction will spontaneous above -419.69 Kelvins.