Question: The equilibrium constant Kc for the equatino CS2(g)+ 4H2(g) goes to CH4(g)+2H2S(g) at 900 degree's C is 27.8. What is the value of Kc for the following equation 1/2CH4(g)+H2S(g)goes to 1/2 CS2(g)+ 2H2(g).
I'm not sure what to do. the only thing i could think of was that it involved maybe the coeficents in some way?
Yes, it's the coefficients.
For example,
If the coefficients are doubled, then K for the new reaction is K^2
If the coefficients are tripled, then K for the new reaction is K^3.
If the coefficients are halved, then K for the new reaction is K^1/2 (sqrt K).
If the reaction is reversed, then K for the new reaction is 1/K.
To find the value of Kc for the given equation, you need to apply the principles of equilibrium constant expressions and stoichiometry.
The given equation is: 1/2CH4(g) + H2S(g) ⇌ 1/2CS2(g) + 2H2(g)
1. Write the balanced chemical equation:
CH4(g) + 2H2S(g) ⇌ CS2(g) + 4H2(g)
2. Determine the equilibrium constant expression (Kc) for this equation. Since there is no coefficient for CH4 and CS2 in the equation, we can assume that their stoichiometric coefficients are both 1:
Kc = ([CS2]^(1/2) x [H2]^4) / ([CH4]^(1/2) x [H2S]^2)
3. Given that the equilibrium constant (Kc) for the reaction CS2(g) + 4H2(g) ⇌ CH4(g) + 2H2S(g) is 27.8, we can use this information to find the value of Kc for the new equation.
To do this, we can use the concept of reverse reaction and inverse the equilibrium constant:
Kc(reverse) = 1 / Kc(forward) = 1 / 27.8
4. Now, since the stoichiometric coefficients in the new equation are halved compared to the forward reaction of the original equation, we adjust the equilibrium constant expression for the new equation accordingly:
Kc(new) = (Kc(reverse))^2 = (1 / 27.8)^2 = 1 / 772.84
Therefore, the value of Kc for the equation 1/2CH4(g) + H2S(g) ⇌ 1/2CS2(g) + 2H2(g) is approximately 0.0013.
Remember, when adjusting equilibrium constants due to changes in stoichiometry, you need to raise or lower the expression to the power of the coefficient changes. In this case, we squared it because the coefficients were halved.