The electrochemical cell described by the balanced chemical reaction has a standard emf (electromotive force) of 0.27 V. Calculate the equilibrium constant (Kc) for the reaction at 298 K. Round your answer to 3 significant figures.
H2O2(aq) + 2ClO2(g) → O2(g) + 2HClO2(aq)
i always do this problem wrong because i can never find the right n
and you use -RTlnK=-nFE0cell right?
H2O2(aq) + 2ClO2(g) → O2(g) + 2HClO2(aq)
O2 in H2O2 on the left is -2 for both O atoms. On the right O2 is zero. Loss of 2e.
2Cl on the left are +8 an on the right 2Cl in HClO2 are +6. Gain of 2e.
n must be 2, right?
thank you Dr. Bob222, i don't know how you manage to answer everyone's questions but you are one amazing person :)
To calculate the equilibrium constant (Kc) for the given reaction at 298 K using the equation -RTlnK = -nFE°cell, you need to determine the value of the stoichiometric coefficient (n) for the reaction.
To find the correct value for n, consider the stoichiometry of the balanced chemical equation. The coefficient in front of each species represents the number of moles of that species involved in the reaction.
In this case, the balanced equation is:
H2O2(aq) + 2ClO2(g) → O2(g) + 2HClO2(aq)
From the equation, you can see that for every 2 moles of ClO2 consumed, 2 moles of HClO2 are produced. Therefore, the value of n for this reaction is 2.
Now, plug the values into the equation -RTlnK = -nFE°cell:
-RTlnK = -2FE°cell
Given that the standard emf (E°cell) is 0.27 V, the Faraday's constant (F) is 96485 C/mol, and the molar gas constant (R) is 8.314 J/(mol·K), you can solve for lnK:
-8.314 J/(mol·K) * (298 K) * ln(K) = -2 * 96485 C/mol * (0.27 V)
Simplifying the equation, you get:
ln(K) = 5.568 / (J/(mol·K))
Now, to solve for K, you need to convert the natural logarithm (ln) to the base 10 logarithm by using the relationship:
ln(K) = log(K) / log(e)
Where log(e) is the natural logarithm of 10: log(e) ≈ 2.303.
Therefore, the equation becomes:
5.568 / (J/(mol·K)) = log(K) / 2.303
Finally, to solve for K, rearrange the equation:
log(K) = 5.568 / (J/(mol·K)) * 2.303
Now, evaluate the right-hand side of the equation and calculate K by taking the antilogarithm (10^x) of both sides:
K = 10^(5.568 / (J/(mol·K)) * 2.303)
Round your answer to 3 significant figures to obtain the equilibrium constant (Kc) for the reaction at 298 K.