True or false? If false, please state why. Every species that appears in the rat

Question

True or false? If false, please state why. Every species that appears in the rate law of a reaction must be a reactant or product in that reaction. Catalysts change rate constant and equilibrium constant of a given reaction. The rate law for the reaction A + B rightarrow product is u = k [A] [B] When an overall reaction is at equilibrium, the forward rate of the reaction must equal the reverse rate of the reaction. The Faraday constant equals the charge per mole of electrons. The gas-phase reaction 2 NO_2 + O_3 rightarrow N_2 O_3 + O_2 has the rate constant k = 2.0 times 10^4 L-mol^-1. s^-1 at 300 K. What is the order of this reaction? The first-order reaction 2A rightarrow 2B + C is 35% complete after 325 sec. Find rate constant k for this reaction. How long will it take for the reaction to be 70% complete? Use the following standard-state cell potentials to calculate the solubility of Mg(OH)_2 at 25 degree C. Mg(OH)_2 + 2 e Mg + 2 OH phi degree = - 2.69 V Mg^2 + 2 e Mg phi degree = - 2.375 V Consider the cell: Cd(s)|CdCl_2 (aq, 0.100 mol/kg) |AgCl(s)| Ag(s) at 25 degree C and 1 bar, in which y_z = 0.228 for the aqueous CdCl_2 solution. Write down the half reactions (indicating anode and cathode) and the overall cell reaction. Find E for the cell given that E degree = 0.624 V. Calculate delta_t G for the cell reaction.

Explanation / Answer

(1) False ; Some times rate limiting step may contain the species which is not in the reaction

(2) False, Catalyst can change rate constant but equilibrium constant will be same.

(3) True, when the reaction is elementary

(4) True, net reaction = forward - backward . At equilibrium net reaction =0

(5) False. 1 Faraday = 1 mol of electrons x Charge of electron.

(2)

From the units of the Rate conatant the order of the reaction can be determined.

K = ( L/mol ) n-1 / s

Here,

n-1 =1

n=2

So the order is 2.

pls post remaining questions separately. According to chegg rules only one question should be answered

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