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).
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Answer:
The correct answer is A. 140 atm
Explanation:
We use the gas formula, which results from the combination of the Boyle, Charles and Gay-Lussac laws. According to which at a constant mass, temperature, pressure and volume vary, keeping constant PV / T. We convert the unit Celsius into Kelvin:
0 ° C = 273K, 67 ° C = 273 + 67 = 340K; 94 ° C = 273 + 94 = 367K
P1xV1 /T1= P2x V2/T2
P2= ((P1xV1 /T1)xT2)/V2
P2=((88,89atm x 17L/340K)x367K)/12L= <em>135,927625 atm</em>
<u>Answer:</u> The standard free energy change of formation of 
is 92.094 kJ/mol
<u>Explanation:</u>
We are given:
Relation between standard Gibbs free energy and equilibrium constant follows:
where,
= standard Gibbs free energy = ?
R = Gas constant = 
T = temperature = ![25^oC=[273+25]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B273%2B25%5DK%3D298K)
K = equilibrium constant or solubility product = 
Putting values in above equation, we get:
For the given chemical equation:
The equation used to calculate Gibbs free change is of a reaction is:
![\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28reactant%29%7D%5D)
The equation for the Gibbs free energy change of the above reaction is:
![\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28Ag%5E%2B%28aq.%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28S%5E%7B2-%7D%28aq.%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28Ag_2S%28s%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol](https://tex.z-dn.net/?f=285.794%3D%5B%282%5Ctimes%2077.1%29%2B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28S%5E%7B2-%7D%28aq.%29%29%7D%29%5D-%5B%281%5Ctimes%20%28-39.5%29%29%5D%5C%5C%5C%5C%5CDelta%20G%5Eo_f_%7B%28S%5E%7B2-%7D%28aq.%29%29%3D92.094J%2Fmol)
Hence, the standard free energy change of formation of is 92.094 kJ/mol
Answer:
Explanation:
CH₃Br+NaOH⟶CH₃OH+NaBr
It is a single step bimolecular reaction so order of reaction is 2 , one for CH₃Br and one for NaOH .
rate of reaction = k x [CH₃Br] [ NaOH]
.008 = k x .12 x .12
k = .55555
when concentration of CH₃Br is doubled
rate of reaction = .555555 x [.24] [ .12 ]
= .016 M/s
when concentration of NaOH is halved
rate of reaction = .555555 x [.12] [ .06 ]
= .004 M/s
when concentration of both CH₃Br and Na OH is made 5 times
rate of reaction = .555555 x .6 x .6
= 0.2 M/s
12) Ethylene glycol and <span>Diphenylamine
</span>
<span>attached images
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hope this helps!
