Consider the reaction: co2(g) + ccl4(g) ⇌ 2 cocl2(g) δg° = 46.9 kj under the following conditions at 25 oc: latex: p_{co_2}p c o 2= 0.459 atm, latex: p_{ccl_4}p c c l 4= 0.984 atm, and latex: p_{cocl_2}p c o c l 2= 0.653 atm, δg for the reaction is , and the forward the reaction is

Answer : The value of is,

Explanation :

The formula used for is:

  ............(1)

where,

= Gibbs free energy for the reaction

=  standard Gibbs free energy  = 46.9 kJ

R = gas constant = 8.314 J/mole.K

T = temperature =

= equilibrium constant

First we have to calculate the value of .

The given balanced chemical reaction is,

The expression for reaction quotient will be :

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Now we have to calculate the value of by using relation (1).

Now put all the given values in this formula, we get:

Therefore, the value of is,

false

because it does not have the potential to be entirely used up

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied