Data:
V1 = 6.7 liter
T1 = 23° = 23 + 273.15 K = 300.15 K
P1 = 0.98 atm
V2 = 2.7 liter
T2 = 125° = 125 + 273.15 K = 398.15 K
P2 = ?
Formula:
Combined law of ideal gases: P1 V1 / T1 = P2 V2 / T2
=> P2 = P1 V1 T2 / (T1 V2)
P2 = 0.98 atm * 6.7 liter * 398.15 K / (300.15K * 2.7 liter)
P2 = 3.22 atm
12+12+6(1)+16 =46 46*.251 = 11.546 gram
A catalyst lowers the activation energy needed to start a reaction
Answer:
0.25 mole of H₂
Explanation:
The equation for the reaction is given below:
Mg + 2HCl —> MgCl₂ + H₂
From the question given above, we were told that the reaction produced 5.6 L of H₂.
Thus, we can obtain obtain the number of mole of H₂ that occupied 5.6 L at STP as follow:
1 mole of H₂ occupied 22.4 L at STP.
Therefore, Xmol of H₂ will occupy 5.6 L at STP i.e
Xmol of H₂ = 5.6 / 22.4
Xmol of H₂ = 0.25 mole
Therefore, 0.25 mole of H₂ were obtained from the reaction.
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