Calculate the standard enthalpy of formation of liquid methanol, ch3oh(l), using the following information: c(graphite) + o2 --> co2(g) latex: \deltaδh° = –393.5 kj/mol h2(g) + (1/2)o2 --> h2o(l) latex: \deltaδh° = –285.8 kj/mol ch3oh(l) + (3/2)o2(g) --> co2(g) + 2h2o(l) latex: \deltaδh° = –726.4 kj/mol

Answer : The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation :

Standard formation of reaction : It is a chemical reaction that forms one mole of a substance from its constituent elements in their standard states.

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of will be,

The intermediate balanced chemical reaction will be,

(1)    

(2)    

(3)    

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, we get :

(1)    

(2)    

(3)    

The expression for enthalpy of formation of will be,

Therefore, the standard enthalpy of formation of methanol is, -238.7 kJ/mole

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(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