Consider the general reaction given below at 400 K x ~A ( ~g ) y ~B ( ~g ) The values of K _ p and K _ c are studied under the same condition of temperature but variation in x and y . (i) K _ p =85.87 and K _ c =2.586 appropriate units (ii) K _ p =0.862 and K _ c =28.62 appropriate units. The values of x and y in (i) and (ii) respectively are:

Step-by-step solution

The relationship between K_p and K_c is given by K_p = K_c(RT)^ n_g , where n_g = y - x . Given T = 400 K and R = 0.0821 L atm K ⁻¹ mol ⁻¹ . The value of RT = 0.0821 400 = 32.84 . For case (i): K_p = 85.87 and K_c = 2.586 . K_p K_c = 85.87 2.586 33.2 . Since 33.2 32.84^1 , we have n_g = y - x = 1 . Checking options for (i): (1) y-x = 3-1 = 2 (2) y-x = 1-4 = -3 (3) y-x = 1-3 = -2 (4) y-x = 2-1 = 1 . This matches. For case (ii): K_p = 0.862 and K_c = 28.62 . K_p K_c = 0.862 28.62 0.0301 . Since 0.0301 1 32.84 = 32.84