Answer:
ΔH = - 5315 kJ.
Explanation:
The given chemical reaction is as follows -
2C₄H₁₀ (g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (g) + 5315 kJ
In the above equation , the amount of energy i.e. 5315 kJ is released , i.e. it is in the product side , hence , the reaction is an example of an exothermic reaction .
Hence ,
The value of the change in enthalphy , i.e. , the enthalpy of product minus the enthalpy of the product .
Therefore ,
The value of the change in enthalphy = - ve .
Hence ,
ΔH = - 5315 kJ.
Here is the correct option: According to Newton's third law of motion, the balloon is pushed forward as the air is forced out. Newton's third law of motion states that for every action there is an equal and opposite reaction. This means that in every interaction, there is a pair of force that is acting on the interacting objects, the size of the force on the first object is equal to the size of the force on the second object.
The mole ratio of the reaction shows that equal volumes of hydrogen gas will be produced by the two reactions.
<h3>What is the mole ratio of a reaction?</h3>
The mole ratio of a reaction is the ratio in which the reactants and products of a given reaction occur for the reaction to proceed to completion.
The mole ratio of a reaction is also known as the stoichiometry of the reaction.
The equation of the two reactions are given below:
From the equation of the reaction reaction, an equal volume of hydrogen gas will be produced by the two reactions.
Therefore, the mole ratio of the reaction shows that equal volumes of hydrogen gas will be produced by the two reactions.
Learn more about mole ratio at: brainly.com/question/19099163
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Answer:
1) ΔG°r(298 K) = - 28.619 KJ/mol
2) ΔG°r will decrease with decreasing temperature
Explanation:
- CO(g) + H2O(g) → H2(g) + CO2(g)
1) ΔG°r = ∑νiΔG°f,i
⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)
from literature, T = 298 K:
∴ ΔG°CO2(g) = - 394.359 KJ/mol
∴ ΔG°CO(g) = - 137.152 KJ/mol
∴ ΔG°H2(g) = 0 KJ/mol........pure substance
∴ ΔG°H2O(g) = - 228.588 KJ/mol
⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )
⇒ ΔG°r(298 K) = - 28.619 KJ/mol
2) K = e∧(-ΔG°/RT)
∴ R = 8.314 E-3 KJ/K.mol
∴ T = 298 K
⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6
⇒ ΔG°r = - RTLnK
If T (↓) ⇒ ΔG°r (↓)
assuming T = 200 K
⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)
⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol
