Question

Consider the following equilibrium reaction:2NO(g) + Br2(g) ⇌ 2NOBr(g)If the initial concentration of NO is 0.10 M, Br2 is 0.20 M, and NOBr is 0.40 M, what would be the concentration of NOBr when additional Br2 is added such that its concentration is doubled? Calculate the new equilibrium concentration of NOBr assuming all other conditions remain the same.

Answer 1

First, we need to find the equilibrium constant  Kc  for the reaction using the initial concentrations:Kc = [NOBr]^2 /  [NO]^2 * [Br2] Using the initial concentrations:Kc =  0.40 ^2 /   0.10 ^2 *  0.20   = 0.16 / 0.002 = 80Now, let's consider the new situation where the concentration of Br2 is doubled, i.e., 0.20 M * 2 = 0.40 M. Let x be the change in concentration of NOBr at the new equilibrium.For the new equilibrium concentrations, we have:[NOBr] = 0.40 + x[NO] = 0.10 - 0.5x  since 2 moles of NO react for every 1 mole of Br2 [Br2] = 0.40 - xNow, we can use the Kc value to find the new equilibrium concentrations:80 = [ 0.40 + x ^2] /  [ 0.10 - 0.5x ^2] *  0.40 - x  Now, we need to solve this equation for x:80 0.10 - 0.5x ^2 0.40 - x  =  0.40 + x ^2This equation is quite complex to solve analytically, so we can use numerical methods or a calculator to find the value of x:x  0.131Now, we can find the new equilibrium concentration of NOBr:[NOBr] = 0.40 + x = 0.40 + 0.131  0.531 MSo, the new equilibrium concentration of NOBr is approximately 0.531 M.