More energy can be absorbed by a larger sample of a substance due to its higher mass.

At 2000 ∘C the equilibrium constant for the reaction 2NO(g)⇌N2(g)+O2(g) is Kc=2.4×103. If the initial concentration of NO is 0.1

Rasek [7]

Answer:

$[N_2]=0.0866M$

Explanation:

Hello there!

In this case, in agreement to the chemical reaction, it is possible for us to figure out the equilibrium concentration of the N2 product, via an ICE table plugged in the equilibrium expression:

$Kc=\frac{[N_2][O_2]}{[NO]^2}\\\\2.4x10^3=\frac{x*x}{(0.175-2x)^2}$

In such a way, when solving for x via quadratic equation or just a solver, it is possible to obtain:

$x_1=0.0866M\\\\x_2=0.0884M$

In such a way, since the root 0.0884 M produce a negative concentration of NO (0.175-2*0.0884=-0.0018M), we infer that the correct root is 0.0866 M; therefore, the concentration of N2 at equilibrium is equal to x:

$[N_2]=x=0.0866M$

Best regards!

4 0

3 years ago

1. If you have a sample of gas at a pressure of 16 atm, what will the pressure be if the volume is halved?

vagabundo [1.1K]

Answer:

1. The pressure will be 32 atm, twice the initial pressure.

2. The pressure will be 1.83 atm, one third of the initial pressure.

Explanation:

Boyle's law is one of the gas laws that relates the volume and pressure of a certain quantity of gas kept at a constant temperature.

This law says that "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

or P * V = k

Ahora es posible suponer que tienes un cierto volumen de gas V1 que se encuentra a una presión P1 al comienzo del experimento. Si varias el volumen de gas hasta un nuevo valor V2, entonces la presión cambiará a P2, y se cumplirá:

P1*V1=P2*V2

1. In this case: