Molecular hydrogen can be made from methane gas by the following reaction: CH_4(

Question

Molecular hydrogen can be made from methane gas by the following reaction: CH_4(g) + H_2O(g) rightarrow CO_2(g) + 3 H_2(g) If H_2 forms at a rate of 2.0 x 10^-3 M/s, at what rate is CH_4 consumed? HCl is slowly added to a solution containing 0.01 M Pb^2+ and 0.01 M Ag^+. Given K_sp (PbCl_2) = 2 4 x 10^-4 and K_sp (AgCl) = 1.6 x 10^-10, which compound precipitates fïrst? Acetic acid (CH_3COOH) is a weak acid (K_a = 1.8 x 10^-5). Calculate the pH of a 17.6 M solution of CH_3COOH (glacial acetic acid). Calculate the equilibrium constant at 25DegreeC for the following redox reaction:

Explanation / Answer

Answer.2 The compound with a lower ksp will precipitate first. So its AgCl

Answer. 3

Equation to calculate pH of a buffer - Henderson Hasselbalch equation:
pH = pKa + log ([A-] /[HA]
Where [A-] = molar concentartion of conjugate base - [CH3COO-] = 17.6M
[HA] Molar concentration of weak acid - [CH3COOH] = 17.6 M
pKa = -logKa = -log 1.8 x 10 -5 /17.6= 0.0178

Log 0.0178 =1.75

Answer. 4 To solve this problem, we first need to know two equations and then recognize the need to combine them to solve for the equilibrium constant:

G = -RTlnK
G = -nFE

Now, set these equations equal to each other:

G = -nFE = -RTlnK

Then rearrange the equation to solve for K:

-nFE = -RTlnK
lnK = nFE/RT
K = e^(nFE/RT)

Now we need to determine the values of the variables above:

n = number of moles of electrons transferred. This is equal to one, since one mole of electrons is transferred from zinc (which is oxidized from an oxidation state of +3 to one of +4) to cadmium (which is reduced from an oxidation state of +2 to one of +1).

E = the net voltage of the reaction.zinc is being oxidized, so we need to change the sign of the standard reduction potential of zinc and add it to the standard reduction potential of cadmium:

E = +0.403 V + (-0.762 V)
E = +0.332V

T = temperature in Kelvins. Since we are given the temperature in degrees Celcius, we need to convert it to Kelvins. Recall that 1 degree C = 1 K and that 0 degrees C = 273 K:

T = 273 + 25
T = 298 K

Now we combine these values with our two constants (F = 96485.34 C/mol and R = 8.3145 J/mol*K) to solve for K:

K = e^(nFE/RT)
K = e^((1)(96485.34 C/mol)(0.332 V) / (8.3145 J/mol*K)(298 K))
K = 2 x 0.332 / 0.059
K = 11.2
K = 10 11.2

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