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Elza [17]
2 years ago
15

The free energy of formation of nitric oxide, NO, at 1000 K (roughly the temperature in an automobile engine during ignition) is

about 78 kJ/mol. Calculate the equilibrium constant Kp for the reaction N2(g) + O2(g) 2NO(g) at this temperature.
Chemistry
1 answer:
Assoli18 [71]2 years ago
4 0

<u>Answer:</u> The value of K_p for the chemical equation is 8.341\times 10^{-5}

<u>Explanation:</u>

For the given chemical equation:

N_2(g)+O_2(g)\rightarrow 2NO(g)

To calculate the K_p for given value of Gibbs free energy, we use the relation:

\Delta G=-RT\ln K_p

where,

\Delta G = Gibbs free energy = 78 kJ/mol = 78000 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant = 8.314J/K mol

T = temperature = 1000 K

K_p = equilibrium constant in terms of partial pressure = ?

Putting values in above equation, we get:

78000J/mol=-(8.314J/Kmol)\times 1000K\times \ln K_p\\\\Kp=8.341\times 10^{-5}

Hence, the value of K_p for the chemical equation is 8.341\times 10^{-5}

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What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig
Mashutka [201]

Answer:

P_H =2.86

c=1.4\times 10^{-4}

Explanation:

first write the equilibrium equaion ,

C_3H_6O_3  ⇄ C_3H_5O_3^{-}  +H^{+}

assuming degree of dissociation \alpha =1/10;

and initial concentraion of C_3H_6O_3 =c;

At equlibrium ;

concentration of C_3H_6O_3 = c-c\alpha

[C_3H_5O_3^{-}  ]= c\alpha

[H^{+}] = c\alpha

K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}

\alpha is very small so 1-\alpha can be neglected

and equation is;

K_a = {c\alpha \times \alpha}

[H^{+}] = c\alpha = \frac{K_a}{\alpha}

P_H =- log[H^{+} ]

P_H =-logK_a + log\alpha

K_a =1.38\times10^{-4}

\alpha = \frac{1}{10}

P_H= 3.86-1

P_H =2.86

composiion ;

c=\frac{1}{\alpha} \times [H^{+}]

[H^{+}] =antilog(-P_H)

[H^{+} ] =0.0014

c=0.0014\times \frac{1}{10}

c=1.4\times 10^{-4}

6 0
2 years ago
In the chemical equation 3c2h4 how many atoms of carbon are represented
Scorpion4ik [409]
An example.
water is H2O

2 hydrogen, 1 oxygen

so the number to the right means how much of what is on the left.

so it looks like 2, because C2, but look at the 3 at the beginning. that means
3 (c2h4)

so 6 carbons, 12 hydrogen

the ratio of c2 to h4 doesn't change it's always 1:2.

but the 3 at the front is a different number relating to how much you have
5 0
2 years ago
Insoluble sulfide compounds are generally black in color. Which of the following combinations could yield a black precipitate? C
Lubov Fominskaja [6]

Answer:

Li₂S(aq) + Pb(NO₃)₂(aq)

2 K₂S(aq) + Sn(NO₃)₄(aq)

Explanation:

For these reactions, the products are:

1. Na₂S(aq) + 2 KCl(aq) → K₂S + 2 NaCl

2. Li₂S(aq) + Pb(NO₃)₂(aq) →  PbS + 2 Li(NO₃)

3. Pb(ClO₃)₂(aq) + 2 NaNO₃(aq) → Pb(NO₃)₂ + 2 Na(ClO₃)

4. AgNO₃(aq) + KCl(aq) → AgCl + KNO₃

5. 2 K₂S(aq) + Sn(NO₃)₄(aq) → SnS₂ + 4 KNO₃

K₂S is a soluble sulfide that, in solid state, is white. As K has a low electronegativity, the relative polar bond K-S allows its dissolution in water.

PbS is a black and insoluble compound. As Pb electronegativity is higher than K electronegativity, The bond is less-polar and its dissolution will not be allowed.

For the third reaction, all nitrates are soluble and Na(ClO₃) is also a very soluble compound.

AgCl is an insoluble white compound.

As Sn electronegativity is high, the solubility of this compound is very low. Also, this compound is black!

Thus, combinations could yield a black precipitate are:

Li₂S(aq) + Pb(NO₃)₂(aq)

2 K₂S(aq) + Sn(NO₃)₄(aq)

I hope it helps!

6 0
2 years ago
An evaporation–crystallization process of the type described in Example 4.5-2 is used to obtain solid potassium sulfate from an
timofeeve [1]

Answer:

feed = 220.77 kg/s; maximum production rate of solid crystal = 416 kg/s; the rate of supplying fresh feed to obtain the production rate = 1.6

Explanation:

Material or mass balance can be used to estimate the mass flow rates of all the streams in the diagram shown in the attached file.

Overall balance: M_{1} = 175 + 11M_{2}

Water: 0.804M_{1} = 175 + 0.6M_{2}

Using substitution method, we have:

M_{1} = 220.77 kg/s

M_{2}  = 4.16 kg/s

The maximum production rate of solid crystal is 10M_{2} = 10*4.16 = 416 kg/s

Around evaporator:

0.45M_{5} = 175

M_{5} = 175/0.45 = 389 kg/s

Around the mixing point:

M_{1} + M_{3} = M_{4} + M_{5}

Solid crystal: 0.196M_{1} + 0.4M_{3} = M_{4}

Using the last two equations, we can obtain:

0.804M_{1} + 0.6M_{3} = M_{5}

0.6M_{3} = 389 - 0.804*220.77 = 389 - 177.5 = 211.5

M_{3} = 211.5/0.6 = 352.5 kg/s

The rate of supplying fresh feed to obtain the production rate is:

\frac{M_{3}}{M_{1}} = 352.5/220.77 = 1.6

7 0
3 years ago
In a reverse reaction, products are formed from reactants and reactants are formed from products. when they yield the same amoun
GREYUIT [131]
.......When they yield the same amount , *Reversible reactions* will reach chemical equilibrium........
8 0
2 years ago
Read 2 more answers
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