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pychu [463]
3 years ago
13

If a reaction mixture contains only N2O and NO2 at partial pressures of 1.0 atm each, the reaction will be spontaneous until som

e NO forms in the mixture. What temperature is required to make the reaction spontaneous under standard conditions?

Chemistry
1 answer:
Vesna [10]3 years ago
3 0

Answer:

Temperature required = 923K

Explanation:

The question is incomplete as there are some details that has to be given. details like the values of the standard enthalpies and entropies of the reactants and product as this is needed to calculate the actual value of the standard enthalpies and standard entropies of the reaction. I was able to get those values from literature and then calculated what needs to be calculated.

From there, I was able to use the equation that shows the relationship between, gibb's free energy, enthalpy, entropy and temperature. The necessary mathematical manipulation were done and the values were plugged in to get the temperature required to make the reaction spontaneous.

A few notes on the Gibb's free energy.

The Gibb's free energy also referred to as the gibb's function represented with letter G. it is the amount of useful work obtained from a system at constant temperature and pressure. The standard gibb's free energy on the other hand is a state function represented as Delta-G, as it depends on the initial and final states of the system.

The spontaneity of a reaction is explained by the standard gibb's free energy.

  • If Delta-G = -ve ( the reaction is spontaneous)
  • if Delta -G = +ve ( the reaction is non-spontaneous)
  • if Delta-G = 0 ( the reaction is at equilibrium)

The step by step calculations is done as shown in the attachment.

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Answer:

heat and large amount of surface area

Explanation:

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4 0
3 years ago
Compute the molar enthalpy of combustion of glucose (C6 H12O6 ): C6 H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2 O (g) Given that com
lana66690 [7]

Answer:

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

Explanation:

Step 1: Data given

Mass of glucose = 0.305 grams

Combustion of 0.305 grams causes a raise of 6.30 °C

Calorimeter has a heat capacity of 755 J/°C

Molar mass of glucose = 180.2 g/mol

Step 2: The balanced equation

C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (g)

Step 3:

ΔH = (m * C * ΔT + c(calorimeter) * ΔT)

with m = mass of the solutin = 0.305 grams

with C = heat capacity of water = 4.184 J/g°C

with ΔT = the change in temperature = 6.30 °C

with c(calorimeter) = 755 J/°C

ΔH = 0.305 * 4.184 *6.30 + 755 * 6.30  = 4764.5 J ( negative because it's exothermic)

Step 4: Calculate moles of glucose

Moles glucose = mass glucose / Molar mass glucose

Moles glucose = 0.305 grams / 180.2 g/mol

Moles glucose = 0.00169 moles

Step 5: Calculate molar enthalpy

Molar enthalpy = -4764.5 J / 0.00169 moles

Molar enthalpy = - 2819254.2 J/moles = -2819.3 kJ/moles

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

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Answer:

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Explanation:

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2 years ago
Read 2 more answers
2 HI(g) ⇄ H2(g) + I2(g) Kc = 0.0156 at 400ºC 0.550 moles of HI are placed in a 2.00 L container and the system is allowed to rea
Alex_Xolod [135]

<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 0.0275 M

<u>Explanation:</u>

Molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}

Moles of HI = 0.550 moles

Volume of container = 2.00 L

\text{Initial concentration of HI}=\frac{0.550}{2}=0.275M

For the given chemical equation:

                          2HI(g)\rightleftharpoons H_2(g)+I_2(g)

<u>Initial:</u>                  0.275

<u>At eqllm:</u>           0.275-2x      x         x

The expression of K_c for above equation follows:

K_c=\frac{[H_2][I_2]}{[HI]^2}

We are given:

K_c=0.0156

Putting values in above expression, we get:

0.0156=\frac{x\times x}{(0.275-2x)^2}\\\\x=-0.0458,0.0275

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.0275 M

Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M

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3 years ago
Write the molecular formula for a compound with the possible elements C, H, N and O that exhibits a molecular ion at M
liberstina [14]

Answer:

= \mathbf{C_3H_6O}

Explanation:

From the given information, since the molecular mass of the ion M+ is not given;

Let's assume M+ = 58.0423

So, by applying the 13th rule;

we will need to divide the mass by 13, after dividing it;

The quotient n = no. of carbon; &

The addition of the quotient (n) with the remainder r =  no. of hydrogen.

So;

\dfrac{58}{13}= 4 \ remainder \ 6

So;

C_nH_{n+r} = C_4H_{4+6}

= C_4H_{10}

From the given information; we have oxygen present, so since the mass of oxygen = 16, we put oxygen in the molecular formula by removing CH_4. Also, since the mass is an even number then Nitrogen is 0.

So, we have:

= \mathbf{C_3H_6O}

6 0
2 years ago
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