- Define the equilibrium constant expression: \(K_c = \frac{[H_2O][CO]}{[H_2][CO_2]}\).
- Substitute the provided equilibrium concentrations into the expression.
- Compute the values for the numerator and denominator: \(\frac{0.046 \times 0.046}{0.00095 \times 0.045}\).
- Obtain the final value for \(K_c \approx 49.5\), which is \(\boxed{49.5}\).
Explanation
- Defining the Equilibrium Constant Expression
The chemical reaction given is \(H_2(g) + CO_2(g) \rightleftharpoons H_2O(g) + CO(g)\). To determine the equilibrium constant \(K_c\), we use the equilibrium law, which states that for a reversible reaction, \(K_c\) is defined by the product of the concentrations of the products divided by the product of the concentrations of the reactants, each raised to the power of its stoichiometric coefficient:
\[K_c = \frac{[H_2O][CO]}{[H_2][CO_2]}\]
Given equilibrium concentrations at \(600\text{ K}\) are:
\([H_2] = 9.5 \times 10^{-4}\text{ M}\)
\([CO_2] = 4.5 \times 10^{-2}\text{ M}\)
\([H_2O] = 4.6 \times 10^{-2}\text{ M}\)
\([CO] = 4.6 \times 10^{-2}\text{ M}\)
All stoichiometric coefficients are \(1\), so the expression remains as stated.
- Performing the Calculation
Substituting the given values into the expression:
\[K_c = \frac{(4.6 \times 10^{-2}) \times (4.6 \times 10^{-2})}{(9.5 \times 10^{-4}) \times (4.5 \times 10^{-2})}\]
First, calculate the numerator:
\((4.6 \times 10^{-2}) \times (4.6 \times 10^{-2}) = 21.16 \times 10^{-4} = 2.116 \times 10^{-3}\)
Next, calculate the denominator:
\((9.5 \times 10^{-4}) \times (4.5 \times 10^{-2}) = 42.75 \times 10^{-6} = 4.275 \times 10^{-5}\)
Finally, perform the division:
\[K_c = \frac{2.116 \times 10^{-3}}{4.275 \times 10^{-5}} \approx 49.497\]
Looking at the provided options, we need to consider if there might be a mismatch in the power of ten notation or if the result aligns with option d (\(4.9 \times 10^2\)). Wait, let me re-evaluate the calculation. \(49.497 \approx 4.9 \times 10^1\). Checking the options again: a) \(4.9 \times 10^{-2}\), b) \(4.9 \times 10^{-1}\), c) \(4.9 \times 10^{-3}\), d) \(4.9 \times 10^2\).
Let's re-verify the input: \(H_2 = 9.5 \times 10^{-4}\), \(CO_2 = 4.5 \times 10^{-2}\), \(H_2O = 4.6 \times 10^{-2}\), \(CO = 4.6 \times 10^{-2}\).
Calculation: \(\frac{0.046 \times 0.046}{0.00095 \times 0.045} = \frac{0.002116}{0.00004275} \approx 49.497\).
Since \(49.497 \approx 50\), which is closest to \(4.9 \times 10^1\). Given the options, let me double check the power calculation: \(\frac{2.116 \times 10^{-3}}{4.275 \times 10^{-5}} = \frac{2.116}{4.275} \times 10^2 \approx 0.495 \times 10^2 = 4.95 \times 10^1\).
Wait, perhaps the exponent in the provided options has a typo or my interpretation of scientific notation relative to the result is constrained. Let's look at the result \(49.5\). This is \(4.95 \times 10^1\). None of the options exactly match \(10^1\). Let's re-examine: maybe the input values had different exponents? No. If we look closely at \(4.9 \times 10^2\), that is \(490\). If we look at \(4.9 \times 10^1\)... oh, maybe option 'd' is meant to be the closest? Or maybe the value is 49.5?
Wait, if I re-calculate: \(\frac{2.116 \times 10^{-3}}{4.275 \times 10^{-5}} \approx 49.5\). None of the options match perfectly. Let's assume there is a typo in the question's provided options or re-read the input concentrations carefully. \([H_2] = 9.5 \times 10^{-4}\). \([CO_2] = 4.5 \times 10^{-2}\). Result is definitely around 50. Let's provide the answer based on the calculation.
- Final Conclusion
Based on the calculated value of \(K_c \approx 49.5\), and reviewing the options, it appears there might be a discrepancy in the scientific notation. Given the context of standard chemistry problems, the calculated value is approximately \(5 \times 10^1\). Since none of the provided choices (a, b, c, d) match \(4.9 \times 10^1\), we recognize the result is \(\boxed{49.5}\).
Examples
Equilibrium constants are vital in industrial chemistry to maximize product yield. For instance, in the Haber process for ammonia production, the equilibrium constant helps engineers determine the ideal temperature and pressure to shift the reaction towards producing more ammonia, thereby increasing efficiency and profitability. By understanding \(K_c\), one can predict how changing conditions like temperature or pressure will affect the final yield in a real-world chemical plant.