This question is describing the following chemical reaction at equilibrium:
And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:
Thus, by recalling the Van't Hoff's equation, we can write:
Hence, we solve for the enthalpy change as follows:
Finally, we plug in the numbers to obtain:
![\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Cfrac%7B-8.314%5Cfrac%7BJ%7D%7Bmol%2AK%7D%20%2Aln%280.25%2F9%29%7D%7B%5B%5Cfrac%7B1%7D%7B%2875%2B273.15%29K%7D%20-%5Cfrac%7B1%7D%7B%2825%2B273.15%29K%7D%20%5D%20%7D%20%5C%5C%5C%5C%5C%5C%5CDelta%20H%3D4%2C785.1%5Cfrac%7BJ%7D%7Bmol%7D)
Learn more:
Atoms are made of subatomic particles known as protons, neutrons, and electrons.
(Small bonus: in order to mimic the technology of replicators like in star trek, you need to mess with protons, neutrons, and electrons) (pls dont report)
Answer:
1.8 moles of O₂
Explanation:
The balance chemical equation for said double replacement (photosynthesis) reaction is as follow;
6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂
According to balance chemical equation,
6 moles of O₂ are produced by = 6 moles of CO₂
So,
1.8 moles of O₂ will be produced by = X moles of O₂
Solving for X,
X = 1.8 mol × 6 mol / 6 mol
X = 1.8 moles of O₂
Stoichiometric problems in which moles are given and moles or other reactant or product asked are the simplest problems. One should only write the balanced chemical equation and perform above method to find the required moles.
zero because its not going anywhere its stationary
hope this helped ^_^
The balanced chemical equation for the combustion of butane is:
Δ
= Σ
Δ
-Σ
Δ
= ![[{8*(-393.5kJ/mol)}+{10*(-241.82kJ/mol)}]-[{2*(-125.6kJ/mol)}+13*(0 kJ/mol)}]](https://tex.z-dn.net/?f=%5B%7B8%2A%28-393.5kJ%2Fmol%29%7D%2B%7B10%2A%28-241.82kJ%2Fmol%29%7D%5D-%5B%7B2%2A%28-125.6kJ%2Fmol%29%7D%2B13%2A%280%20kJ%2Fmol%29%7D%5D)
=[-3148kJ/mol+(-2418.2kJ/mol)]-[(-251.2kJ/mol)+0]
= -5315 kJ/mol
Calculating the enthalpy of combustion per mole of butane:
Therefore the heat of combustion per one mole butane is -2657.5 kJ/mol
Correct answer: -2657.5 kJ/mol
