Question

A.)Consider the following reaction at 25°C: A(g) + 2B(g) → C(g) Kc= 4.47x106 What is the value of the rate constant in terms of pressure, Kp? Express your answer in scientific notation. KP = KC(RT)Δn (for my answer i got 7475.0133 or 7.48E3 and didnt know if it was right) A(g) + 2B(g) → 2C(g) B.) If Kp for the above reaction is 4.50 x 10-2, what is the value of Q if the pressures of A, B, and C are 0.250 atm, 0.18 atm, and 0.450atm respectively? (i got 25 and didnt know if it was right) A(g) + 2B(g) → 2C(g) C.)If Kp for the above reaction is 4.50 x 10-2, in what direction will the reaction proceed to establish equilibrium if the pressures of A, B, and C are 0.575 atm, 0.850 atm, and 0.015 atm respectively? a)proceeds in the forward direction to establish equilibrium b)proceeds in the reverse direction to establish equilibrium (wouldnt this be the correct answer because the number is much less then one when you do the math?) c)it is at equilibrium already

Answer 1

Part A: The correct value for the rate constant, Kp, is approximately 7.48E3.

Part B: The correct value for Q is 25.

Part C: The reaction will proceed in the reverse direction to establish equilibrium (b).

Step-by-step explanation:

Part A:The expression relating Kp to Kc is given by
`\(K_p = K_c(RT)^Δn\)`, where \(Δn\) is the change in the number of moles of gas between the reactants and products.

In this case, since A(g) + 2B(g) → 2C(g), \(Δn = (1 + 2) - 2 = 1\). Substituting the given values,
`\(K_p = 4.47 × 10^6(RT)^1\)`, and at 25°C
`(\(298.15 K\))`, the correct calculation yields
`\(K_p ≈ 7.48 × 10^3\)`.

Part B: For the reaction
`\(A(g) + 2B(g) → 2C(g)\)`, the expression for Q is given by
`\(Q = [C]^2 / ([A][B]^2)\)`. Substituting the given pressures,
`\(Q = (0.450^2) / (0.250 × 0.18^2) = 25\)`.

Part C: To determine the direction of the reaction, compare Q and Kp. If
`\(Q < K_p\)`, the reaction will proceed in the forward direction; if
`\(Q > K_p\)`, it will proceed in the reverse direction.

In this case, Q = 25 and
`\(K_p = 4.50 × 10^(−2)\)`, thus \(Q > K_p\), and the reaction will proceed in the reverse direction to establish equilibrium (b). This is because the system is currently "too product-rich," and the reaction needs to shift towards the reactants to reach equilibrium.