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Consider the reaction 2N2(g) O2(g)2N2O(g) Using the standard thermodynamic data in the tables linked above, calculate Grxn for t

ratelena [41]

Answer:

$\Delta G^0 _{rxn} = 207.6\ kJ/mol$

ΔG ≅ 199.91 kJ

Explanation:

Consider the reaction:

$2N_{2(g)} + O_{2(g)} \to 2N_2O_{(g)}$

temperature = 298.15K

pressure = 22.20 mmHg

From, The standard Thermodynamic Tables; the following data were obtained

$\Delta G_f^0 \ \ \ N_2O_{(g)} = 103 .8 \ kJ/mol$

$\Delta G_f^0 \ \ \ N_2{(g)} =0 \ kJ/mol$

$\Delta G_f^0 \ \ \ O_2{(g)} =0 \ kJ/mol$

$\Delta G^0 _{rxn} = 2 \times \Delta G_f^0 \ N_2O_{(g)} - ( 2 \times \Delta G_f^0 \ N_2{(g)} + \Delta G_f^0 \ O_{2(g)})$

$\Delta G^0 _{rxn} = 2 \times 103.8 \ kJ/mol - ( 2 \times 0 + 0)$

$\Delta G^0 _{rxn} = 207.6\ kJ/mol$

The equilibrium constant determined from the partial pressure denoted as $K_p$ can be expressed as :

$K_p = \dfrac{(22.20)^2}{(22.20)^2 \times (22.20)}$

$K_p = \dfrac{1}{ (22.20)}$

$K_p$ = 0.045

$\Delta G = \Delta G^0 _{rxn} + RT \ lnK$

where;

R = gas constant = 8.314 × 10⁻³ kJ

$\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15 \ ln(0.045)$

$\Delta G =207.6 + 2.4788191 \times \ ln(0.045)$

$\Delta G =207.6+ (-7.687048037)$

$\Delta G =$ 199.912952 kJ

ΔG ≅ 199.91 kJ

7 0

3 years ago

What is true of a solution of 1.0M HCL

Charra [1.4K]

Answer:

What is true of a solution of 1.0 M HCl(aq)? It contains 1.0 gram of HCl per 100 grams of water.

Explanation:

8 0

3 years ago

Critical practices to control the amount of solid waste include __________.

erica [24]

Reduce, reuse, recycle, compost. I think hope its helpful.

3 0

3 years ago

Read 2 more answers

How many liters of ammonia gas can be formed from 54.1 liters of hydrogen

Aleks04 [339]

<span>The reaction is N2 + 3H2 -> 2NH3
So the amount of NH3 formed is 2/3 of the amount of H2 = 2/3 * 13.7 = 9.13 Liters.</span><span>The answer is 9.13

</span>

5 0

3 years ago

A corpse discovered in the desert at 5:00 PM was found to have larvae from the blow fly species Phormia regina that had just dev

horrorfan [7]

Answer:

12.19 hours prior to discovery was the corpse placed in the desert.

Explanation:

Heat required for fly eggs to develop into first instar larvae = $Exposure\,time\times temperature$

From the given,

Exposure time = 16 hrs

Temperature = 27 C

$=16\times 27 = 427.2^{o}C$

The day time exposed hours is 10 hrs.-(7.00am-5.00pm)

$=10\times 37.8 = 378^{o}C$

So, the extra heat = 427.2-378= 49.2

The addition heat divided by average temperature.

Average temperature = 22.4 C

$=\frac{49.2}{22.4}= 2.19 hrs$

So, the total time exposed is night time hours to day time hours.

= 10+2.19 = 12.19 hrs.

Therefore, 12.19 hours prior to discovery was the corpse placed in the desert.

3 0

3 years ago