Question
how do you write the equilibrium expression for: C3H8 + SO2-><- 3CO2 + 4H2O?
Answers
DrBob222
The equilibrium expression is the product of the products divided by the product of the reactants each raised to a power indicated by the coefficient in the balanced chemical equation. HOWEVER, you must have mixed up two equations because there is no way this will work. I might buy it if the S of the SO2 was not there. I would be willing to bet money that you typed an S (for SO2) but you intended to type 5O2. Please clarify.
madison
i don't know what it is but that's the equation on my hw
madison
im sorry your right it is a 5 :)
DrBob222
how do you write the equilibrium expression for:
C3H8 + 5O2-><- 3CO2 + 4H2O
Keq = (CO2)<sup>3</sup>(H2O)<sup>4</sup>/(C3H8)(O2)<sup>2</sup>.
I am assuming, since this is a combustion reaction, that the H2O is present as a gas. If it isn't, sometimes the water doesn't enter into the equilibrium OR it is essentially constant, in which case it is often omitted from the expression.
C3H8 + 5O2-><- 3CO2 + 4H2O
Keq = (CO2)<sup>3</sup>(H2O)<sup>4</sup>/(C3H8)(O2)<sup>2</sup>.
I am assuming, since this is a combustion reaction, that the H2O is present as a gas. If it isn't, sometimes the water doesn't enter into the equilibrium OR it is essentially constant, in which case it is often omitted from the expression.
madison
thank you
DrBob222
Madison--I made a typo, also. The power of O2 should be to the fifth and not squared. Each concentration is raised to the power of the coefficient in the equation. The coefficient for O2 is 5, therefore (O2)^5. Sorry bout that.
DrBob222
See my last response just above your last one. It points out a typo I made.