Kp/Kc = RT
<h3>Further explanation</h3>
Given
Reaction
A(g) ⇌ C(g)+B(g)
Required
Kp/Kc
Solution
For reaction :
pA + qB ⇒ mC + nD
![\large {\boxed {\bold {Kc ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}](https://tex.z-dn.net/?f=%5Clarge%20%7B%5Cboxed%20%7B%5Cbold%20%7BKc%20~%20%3D%20~%20%5Cfrac%20%7B%5BC%5D%20%5E%20m%20%5BD%5D%20%5E%20n%7D%20%7B%5BA%5D%20%5E%20p%20%5BB%5D%20%5E%20q%7D%7D%7D%7D)
While the equilibrium constant Kp is based on the partial pressure
![\large {\boxed {\bold {Kp ~ = ~ \frac {[pC] ^ m [pD] ^ n} {[pA] ^ p [pB] ^ q}}}}](https://tex.z-dn.net/?f=%5Clarge%20%7B%5Cboxed%20%7B%5Cbold%20%7BKp%20~%20%3D%20~%20%5Cfrac%20%7B%5BpC%5D%20%5E%20m%20%5BpD%5D%20%5E%20n%7D%20%7B%5BpA%5D%20%5E%20p%20%5BpB%5D%20%5E%20q%7D%7D%7D%7D)
The value of Kp and Kc can be linked to the formula '
R = gas constant = 0.0821 L.atm / mol.K
Δn=moles products - moles reactants or
number of product coefficients-number of reactant coefficients
For reaction :
A(g) ⇌ C(g)+B(g)
number of product coefficients = 1+1=2
number of reactant coefficients = 1
Δn= 2 - 1 =1
So Kp/Kc = RT
Answer:
The fuel value of 1 gram of this carbohydrate in a nutritional calorie is 4.0 kcal/G.
Answer:
Electrons are particles that surround the nucleus of an atom like a cloud. As with protons and neutrons, electrons are essential to an atom's structure.
M.P of oleic acid = 13 deg C
<span>B.Pof oleic acid = 360 deg C </span>
<span>(360 - 13) / 100 = 3.47 </span>
<span>So 1 deg O = 3.47 deg C </span>
<span>The scale do not start at 0 deg C, it starts at 13 deg C. So to convert deg C to deg O,
subtract 13 then divide by 3.47 </span>
<span>deg O = (deg C - 13) / 3.47 </span>
<span>convert O to C multiply by 3.47 then add 13 </span>
<span>deg C = (deg O x 3.47) + 13 </span>
<span>convert 0 deg C to deg O </span>
<span>deg O = (0 deg C - 13) / 3.47 </span>
<span>= - 3.75 deg O</span>
