Answer:
Temperature required = 923K
Explanation:
The question is incomplete as there are some details that has to be given. details like the values of the standard enthalpies and entropies of the reactants and product as this is needed to calculate the actual value of the standard enthalpies and standard entropies of the reaction. I was able to get those values from literature and then calculated what needs to be calculated.
From there, I was able to use the equation that shows the relationship between, gibb's free energy, enthalpy, entropy and temperature. The necessary mathematical manipulation were done and the values were plugged in to get the temperature required to make the reaction spontaneous.
A few notes on the Gibb's free energy.
The Gibb's free energy also referred to as the gibb's function represented with letter G. it is the amount of useful work obtained from a system at constant temperature and pressure. The standard gibb's free energy on the other hand is a state function represented as Delta-G, as it depends on the initial and final states of the system.
The spontaneity of a reaction is explained by the standard gibb's free energy.
- If Delta-G = -ve ( the reaction is spontaneous)
- if Delta -G = +ve ( the reaction is non-spontaneous)
- if Delta-G = 0 ( the reaction is at equilibrium)
The step by step calculations is done as shown in the attachment.
Answer:
Octane is a hydrocarbon and an alkane with the chemical formula C 8 H 18, and the condensed structural formula CH 3 (CH 2) 6 CH 3.Octane has many structural isomers that differ by the amount and location of branching in the carbon chain. One of these isomers, 2,2,4-trimethylpentane (commonly called iso-octane) is used as one of the standard values in the octane rating scale.
Chemical formula: C₈H₁₈
Molar mass: 114.232 g·mol−1
Melting point: −57.1 to −56.6 °C; −70.9 to −69.8 °F; 216.0 to 216.6 K
Solubility in water: 0.007 mg dm−3 (at 20 °C)
Answer:
A resource is a source or supply from which a benefit is produced.
Explanation:
hope it helped :)
441 g CaCO₃ would have to be decomposed to produce 247 g of CaO
<h3>Further explanation</h3>
Reaction
Decomposition of CaCO₃
CaCO₃ ⇒ CaO + CO₂
mass CaO = 247 g
mol of CaO(MW=56 g/mol) :
From equation, mol ratio CaCO₃ : CaO = 1 : 1, so mol CaO :
mass CaCO₃(MW=100 g/mol) :