Answer: B 15.3 Algorithmic Questions 2) Given the following reaction at equilibrium, if K_(c)=5.84times 10^5 at 230.0^circ C,K_(p)= 2NO(g)+O_(2)(g)leftharpoons 2NO_(2)(g) A) 3.67times 10-2 B) 1.41times 10^4 C) 6.44times 105 D) 2.40times 106

To find the value of Kp, we can use the relationship between Kc and Kp:<br /><br />Kp = Kc * (RT)^(Δn)<br /><br />where R is the gas constant, T is the temperature in Kelvin, and Δn is the change in the number of moles of gas.<br /><br />First, let's calculate the change in the number of moles of gas (Δn):<br /><br />Δn = (2 moles of NO2) - (2 moles of NO + 1 mole of O2) = 2 - 3 = -1<br /><br />Now, we can plug in the values into the equation:<br /><br />Kp = Kc * (RT)^(Δn)<br />Kp = 5.84 x 10^5 * (0.0821 * 513.15)^(-1)<br />Kp = 5.84 x 10^5 * (0.0821 * 513.15)^(-1)<br />Kp = 5.84 x 10^5 * (0.0421)^(-1)<br />Kp = 5.84 x 10^5 * 23.6<br />Kp = 1.39 x^4<br /><br />Therefore, the correct answer is B) $1.41\times 10^{4}$.