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lana66690 [7]
3 years ago
7

Ammonium carbamate (NH2COONH4) is a salt of carbamic acid that is found in the blood and urine of mammals. At 240°C, Kc = 0.006

for the following equilibrium: If 6.01 g of NH2COONH4 is put into a 0.859-L evacuated container, what is the total pressure (in atm) at equilibrium? Enter a number to 2 decimal places.
Chemistry
1 answer:
SashulF [63]3 years ago
8 0

Missing information:

Equilibrium reaction: NH2COONH4(s)⇌2NH3(g)+CO2(g)

Answer:

14.43 atm

Explanation:

The equilibrium occurs when in a reversible reaction, the velocity of the formation of the products is equal to the velocity of the formation of the reactants. When this occurs, the concentration and partial pressures remain constant. To characterize the equilibrium there's the equilibrium constant, which can be based on the concentration (Kc) or based on the gases partial pressure (Kp).

The conversion between them is:

Kp = Kc*(RT)⁻ⁿ

Where n is the variation of the coefficients of the gas substances (reactants - products) so, n = 0 - (2+1) = -3

Kp = Kc*(RT)³

R is the gas constant (0.082 atm.L/mol.K), and T is the temperature ( 240°C = 513 K), so:

Kp = 0.006*(0.082*513)³

Kp = 446.6

The value of Kp is also the ratio between the multiplication of the partial pressures of the products divided by the partial pressure of the reactants, each one elevated by the coefficient of the substance, and only for gas substances, so:

Kp = (pNH3)²*(pCO2)

Doing an equilibrium chart

NH3 CO2

0 0 Intial

+2x x Reacts (stoichiometry is 2:1)

2x x Equilibrium

446.6 = (2x)²*x

4x³ = 446.6

x³ = 111.65

x = ∛111.65

x = 4.81 atm

pNH3 = 2*4.81 = 9.62 atm

pCO2 = 4.81 atm

By Dalton's law, the total pressure of a gas mixture is the sum of the partial pressure of the substances:

P = 9.62 + 4.81 = 14.43 atm

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Whats the voltage of CuCl2 + Zn -&gt; ZnCl2 + Cu
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Answer:

Approximately 1.10\; {\rm V} under standard conditions.

Explanation:

Equation for the overall reaction:

{\rm CuCl_{2}}\, (aq) + {\rm Zn}\, (s) \to {\rm ZnCl_{2}} \, (aq) + {\rm Cu}\, (s).

Write down the ionic equation for this reaction:

\begin{aligned}& {\rm Cu^{2+}}\, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Zn}\, (s)\\ & \to {\rm Zn^{2+}} \, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Cu}\, (s)\end{aligned}.

The net ionic equation for this reaction would be:

{\rm Cu^{2+}}\, (aq) + {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + {\rm Cu}\, (s).

In this reaction:

  • Zinc loses electrons and was oxidized (at the anode): {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}.
  • Copper gains electrons and was reduced (at the cathode): {\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s).

Look up the standard potentials for each half-reaction on a table of standard reduction potentials.

Notice that {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} is oxidation and is likely not on the table of standard reduction potentials. However, the reverse reaction, {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s), is reduction and is likely on the table.

  • E(\text{anode}) = -0.7618\; {\rm V} for {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s), and
  • E(\text{cathode}) = 0.3419\; {\rm V} for {\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s).

The reduction potential of {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} would be -E(\text{anode}) = -(-0.7618\; {\rm V}) = 0.7618\; {\rm V}, the opposite of the reverse reaction {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s).

The standard potential of the overall reaction would be the sum of the standard potentials of the two half-reactions:

\begin{aligned} E^{\circ} &= E^{\circ}(\text{cathode}) + (-E^{\circ}(\text{anode})) \\ &= 0.3419 - (-0.7618\; {\rm V}) \\ &\approx 1.10\; {\rm V}\end{aligned}.

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If we have 1.23 mol of NaOH in solution and 0.85 mol of Cl2 gas is available to react, which one is the limiting reactant? Give
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Answer:

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Explanation:

Hello there!

In this case, since the reaction taking place between sodium hydroxide and chlorine has is:

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Which must be balanced according to the law of conservation of mass:

2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O

Whereas there is a 2:1 mole ratio of NaOH to Cl2, which means that the moles of the former that are consumed by 0.85 moles of the latter are:

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