p = mv. Therefore, and increase in either velocity or mass will lead to an increase in momentum and vice versa. Momentum during a reaction is always conserved, meaning that the mass and initial velocity before a reaction will always be equal to the change in mass and velocity produced after the reaction. Kinetic energy after a reaction, however, is not always conserved. For example if a fast moving vehicle collided with a stationary vehicle, and moved together, the overall kinetic energy would be after the reaction, as a heaver mass would be moved by the same velocity causing a decrease in kinetic energy.

I don't know if this is exactly what you are looking for, but in physics this is how it is understood.

4 0

2 years ago

Be sure to answer all parts.The equilibrium constant, Kc, for the formation of nitrosyl chloride from nitric oxide and chlorine,

djverab [1.8K]

<u>Answer:</u> The reaction proceeds in the forward direction

<u>Explanation:</u>

For the given chemical equation:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

Relation of $K_p\text{ with }K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta n_g}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration = $6.5\times 10^4$

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $35^oC=[35+273]K=308K$

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$K_p=6.5\times 10^4\times (0.0821\times 500)^{-1}\\\\K_p=1583.43$

$K_p$ is the constant of a certain reaction at equilibrium while $Q_p$ is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.