Answer:
When the coefficients in a balanced chemical reaction are multiplied by two, the equilibrium constant is not affected.
Explanation:
The equilibrium constant of a reaction is known to remain steady.
Even if all the coefficients of all the species in the reaction are multiplied by two, the value of the equilibrium constant will reamin the same because the equilibrium position will not change as a result of that.
Answer:
The standard enthalpy of formation of methanol is, -238.7 kJ/mole
Explanation:
The formation reaction of CH_3OH 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, Using Hess's law:
We get :
..[1]
..[2]
[3]
The expression for enthalpy of formation of 
will be,
The standard enthalpy of formation of methanol is, -238.7 kJ/mole
<span>So to make it clear let's break the equation down species by species and assess the number of each species on bothe sides of the equation:
2C</span>₈H₈ + 25O₂ → 8CO₂ + 18H₂<span>O
LHS: C - 16 RHS: C - 8
H - 16 H - 36
O - 50 O - 34
Thus based on that it is evident that the equation is not quite balanced. This therefore means a "</span><span>No, because the number of carbon, hydrogen & oxygen atoms on both sides of the equation are not equal."
</span>The actual balance equation would be C₈H₈ + 10O₂ → 8CO₂ + 4H₂O
