The following reaction is a ______

Questions 18-20 refer to the reaction 2NH3 (g) ↔ N2 (g) + 3H2 (g), and ΔH = 92.2 kJ.

Fittoniya [83]

18. Answer is A

Since the enthalpy of reaction is positive, the forward reaction is an endothermic reaction which means the energy is gained from the surrounding to happen the reaction. If the temperature decreases, according to the Le Chatelier's principle, the system tries to become equilibrium by increasing temperature. Since forward reaction is endothermic (because of the bond breaking), the backward reaction is exothermic (because of the bond making) which releases the energy to the surroundings. This makes the increase of temperature. So if the backward reaction is promoted because of the decrease of temperature, then the concentration of H₂ will decrease.

19. Answer is A.

The reactant side has 2 moles/molecules of reactants and the product side has 4 moles/molecules of products which come from 1 N₂(g) and 3 H₂(g). If the pressure is reduced in the system, according to the Le Chatelier's principle, the system tries to increase the pressure. Hence, forward reaction is promoted because of the higher number of molecules in product side. If the forward reaction is promoted, the concentration of NH₃(g) will decreased.

20. Answer is C.

If the concentration of reactant is increased in the system, according to the Le Chatelier's principle, the system tries to reduce the concentration of that reactant. So if NH₃(g) concentration is increased, then to be equilibrium, the forward reaction will be promoted. Then the concentration of N₂(g) will increase.

4 0

2 years ago

Consider the following reaction where Kc = 154 at 298 K.2NO(g) + Br2(g) 2NOBr(g)A reaction mixture was found to contain 4.64×10-

IgorLugansk [536]

Answer:

The reaction is not at equilibrium and reaction must run in forward direction.

Explanation:

At the given interval, concentration of NO = $\frac{4.64\times 10^{-2}}{1}M=4.64\times 10^{-2}M$

Concentration of $Br_{2}$ = $\frac{4.56\times 10^{-2}}{1}M=4.56\times 10^{-2}M$

Concentration of NOBr = $\frac{0.102}{1}M=0.102M$

Reaction quotient, $Q_{c}$ , for this reaction = $\frac{[NOBr]^{2}}{[NO]^{2}[Br_{2}]}$

species inside third bracket represents concentrations at the given interval.

So, $Q_{c}=\frac{(0.102)^{2}}{(4.64\times 10^{-2})^{2}\times (4.56\times 10^{-2})}=106$

So, the reaction is not at equilibrium.

As $Q_{c}< K_{c}$ therefore reaction must run in forward direction to increase $Q_{c}$ and make it equal to $K_{c}$ .

4 0

2 years ago

Monique hears on the weather report that cold, dry air from Canada is going to hit warm, moist air from the Gulf of Mexico. She

lara [203]

Answer:

She should suspect a Tornado.

Explanation:

Tornadoes form when warm humid air collides with cold dry air. The denser the cold air is pushed over warm air, causing an updraft. As soon as it reaches the ground a tornado is formed.

4 0

2 years ago

Why isn't there a lunar eclipse every time Earth is in between the sun and the Moon?

DerKrebs [107]

Because the Earth's orbit around the sun is not in the same plane as the Moon's orbit around the Earth.

8 0

2 years ago

If the elastic modulus of cobalt (co) is 200 gpa, and the elastic modulus of tungsten carbide (wc) is 700 gpa, calculate the upp

Dmitry_Shevchenko [17]

To determine the upper bond
Ec(u) = EmVm + EpVp
Em is the elastic modulus of cobalt.
E₁ is the elastic modulus of the particulate
Vm is the volume fraction of cobalt
Vp is the volume fraction of particulate
substitute
Ec(u) = 200 (Vm) + 700 (Vp)
To determine the lower bound
Ec (l) = EmEp/VmEp+ VpEm
Substitute
Ec (l) = 200(700)/Vm(700) + Vp (200)
Ec (l) = 1400/7Vm+2Vp

7 0

3 years ago

The following reaction is a _______ reaction.