For X + Y → XY, if [XY] is increasing at a rate 0.10 M s-1, predict the rate of depletion of both reactants.

Calcule a variação da entalpia dessa reação ( 2 NH3 (g) ---> CO(NH2)2 (s) + H2O (L) ) a partir das seguintes equações termoqu

Nitella [24]

ΔH = +438 kJ

We have three equations:

(I) N₂ + 3H₂ → 2NH₃; ΔH = -92 kJ

(II) H₂ +½O₂ → H₂O; ΔH = -286 kJ

(III) CO(NH₂)₂ + ³/₂O₂ → CO₂ + 2H₂O + N₂; ΔH = -632 kJ

From these, we must devise the target equation:

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; ΔH = ?

_________________________________

The target equation has 2NH₃ on the left, so you reverse equation (I).

When you reverse an equation, you reverse the sign of its ΔH.

(V) 2NH₃ → N₂ + 3H₂; ΔH = +92 kJ

Equation (V) has 1N₂ on the right, and that is not in the target equation.

You need an equation with 1N₂ on the left.

Reverse Equation (III).

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; ΔH = +632 kJ

Equation (VI) has ³/₂O₂ on the right, and that is not in the target equation.

You need ³/₂O₂ on the left.

Multiply Equation (II) by three.

When you multiply an equation by three, you multiply its ΔH by three.

(VII) 3H₂ +³/₂O₂ → 3H₂O; ΔH = -286 kJ

Now, you add equations (V), (VI), and (VII), cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

_______________________________________

We get the target equation (IV):

(V) 2NH₃ → N₂ + 3H₂; ΔH = + 92 kJ

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; ΔH = +632 kJ

(VII) 3H₂ +³/₂O₂ → 3H₂O; ΔH = -286 kJ

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; ΔH = +438 kJ

7 0

2 years ago

What are the protons neutrons and electrons of Vanadium-52 +3 charge HELP

Alex17521 [72]

Answer:

Protons: 23

Neutrons: 28

Electrons: 23

I THINK

Explanation:

4 0

3 years ago

How many atoms of oxygen is in CO2?

rosijanka [135]

Answer: 2 atoms

Explanation: 2 in Co2 means 2 atoms

7 0

3 years ago

How much is 2.50 g of CuCl2 in moles ?

Inessa05 [86]

The molar mass of CuCl2 is 134.45 g/mol; therefore, you divide 2.5 g of CuCl2 by 134.45 g of CuCl2 leaving you with 0.019 moles.
I hope this works.
PLEASE GIVE ME A BRAINIEST CROWN.

5 0

3 years ago

How many grams of propane are in 20 pounds of propane? Use the conversion 1 lb = 454 g. (Express your answers for the next three

Dima020 [189]

Answer:

a. $=9.1x10^3gC_3H_8$

b. $2.1x10^2molC_3H_8$

c. $Q=-4.6x10^5kJ$

Explanation:

Hello,

a. By applying the given information, one obtains:

$20lbC_3H_8*\frac{454gC_3H_8}{1lbC_3H_8} =9.1x10^3gC_3H_8$

b. By knowing that the propane has a molecular mass of 44g/mol, one obtains:

$20lbC_3H_8*\frac{454gC_3H_8}{1lbC_3H_8}*\frac{1molC_3H_8}{44gC_3H_8} =2.1x10^2molC_3H_8$

c. Here, the propane's combustion chemical reaction is stated:

$C_3H_8+5O_2-->3CO_2+4H_2O$

This enthalpy of reaction is computed via:

Δ $rH=(3*-393.5kJ/mol)+(4*241kJ/mol)-(-104.7kJ/mol)=-2043kJ/mol$

Finally, since it is done for 20 lb of propane ( $2.1x10^2molC_3H_8$ ), the obtainable energy is:

$Q=-2043kJ/mol*2.1x10^2molC_3H_8\\Q=-4.6x10^5kJ$

Best regards.

4 0

3 years ago