Answer:

Explanation:
We know, 
where, R = 0.0821 L.atm/(mol.K), T is temperature in kelvin and
is difference in sum of stoichiometric coefficient of products and reactants
Here
and T = 311 K
So, ![K_{p}=(0.0111)\times [(0.0821L.atm.mol^{-1}.K^{-1})\times 311K]^{-1}=4.35\times 10^{-4}](https://tex.z-dn.net/?f=K_%7Bp%7D%3D%280.0111%29%5Ctimes%20%5B%280.0821L.atm.mol%5E%7B-1%7D.K%5E%7B-1%7D%29%5Ctimes%20311K%5D%5E%7B-1%7D%3D4.35%5Ctimes%2010%5E%7B-4%7D)
Hence value of equilibrium constant in terms of partial pressure
is 
Explanation:
substance Q could be <em><u>oxygen (O2)</u></em>
substance R could be <em><u>carbon</u></em><em><u> </u></em><em><u>d</u></em><em><u>i</u></em><em><u>o</u></em><em><u>x</u></em><em><u>i</u></em><em><u>d</u></em><em><u>e</u></em><em><u> </u></em><em><u>(</u></em><em><u>C</u></em><em><u>O</u></em><em><u>2</u></em><em><u>)</u></em>
Since the half-reaction is occurring in a basic solution, add 32OH− to each side of the equation to eliminate the H+ ions.
P₄ +16H₂O + 32OH⁻ ⟶ 4PO₃⁻⁴ + 32H⁺ +32OH⁻
Final reaction :
P₄ + 32OH⁻ ⟶ 4PO₃⁻⁴ + 16H₂O + 20e⁻
A half reaction is either the oxidation or reduction reaction component of a redox reaction. A half reaction is obtained by considering the change in oxidation states of individual substances involved in the redox reaction.
The concept of half-reactions is used to describe what occurs in an electrochemical cell, such as a Galvanic cell battery. Half-reactions can be written to describe both the metal undergoing oxidation (known as the anode) and the metal undergoing reduction (known as the cathode).
Half-reactions are often used as a method of balancing redox reactions. For oxidation-reduction reactions in acidic conditions, after balancing the atoms and oxidation numbers, one will need to add H+ ions to balance the hydrogen ions in the half reaction.
For oxidation-reduction reactions in basic conditions, after balancing the atoms and oxidation numbers, first treat it as an acidic solution and then add OH- ions to balance the H+ ions in the half reactions (which would give H2O).
Learn more about Half reactions here : brainly.com/question/2491738
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I think maybe reptiles bu i could be wrong. :)