From the calculations performed, the free energy change for the reaction is 72 kJ/mol.
<h3>What is the equilibrium constant?</h3>
The equilibrium constant is a value that shows the extent to which reactants have been converted to products.
Given that the equation of the reaction is;
3CH4(g)→C3H8(g)+2H2(g)
Then;
PC3H8 = 0.013 atm
PH2 = 2.3×10−2 atm
PCH4 = 41 atm
Now;
ΔG = ΔG° + RTlnQ
ΔG°reaction = ΔG°products - ΔG°reactants
ΔG°reaction = [( -23.4) +2(0)] - 3(-50.8)
ΔG°reaction = 129 kJ/mol
Q = PC3H8 * PH2^2/PCH4^3
Q = 0.013 * (2.3×10−2)^2/( 41)^3
Q = 6.877 * 10^-6/68921
Q= 9.9* 10^-11
Hence;
ΔG = 129 * 10^3 + [8.314 * 298 * (ln 9.9* 10^-11 )]
ΔG = 129 * 10^3 - 57073
ΔG = 72 kJ/mol
Learn more about free energy change: brainly.com/question/14143095
To solve this problem, we can simply calculate for the
dose by multiplying the volume of solution containing Selenium 75 and the
activity of the Selenium 75. That is:
dose = 4.1 mL * (45 μCi/mL)
dose = 184.5 μCi
Answer:
About 0.1738 liters
Explanation:
Using the formula PV=nRT, where p represents pressure in atmospheres, v represents volume in liters, n represents the number of moles of ideal gas, R represents the ideal gas constant, and T represents the temperature in kelvin, you can solve this problem. But first, you need to convert to the proper units. 215ml=0.215L, 86.4kPa is about 0.8527 atmospheres, and 15C is 288K. Plugging this into the equation, you get:

Now that you know the number of moles of gas, you can plug back into the equation with STP conditions:

Hope this helps!
Answer:
21.86582KJ
Explanation:
The graphical form of the Arrhenius equation is shown on the image attached. Remember that in the Arrhenius equation, we plot the rate constant against the inverse of temperature. The slope of this graph is the activation energy and its y intercept is the frequency factor.
Applying the equation if a straight line, y=mx +c, and comparing the given equation with the graphical form of the Arrhenius equation shown in the image attached, we obtain the activation energy of the reaction as shown.