An industrial process involves the reaction of nitrogen gas and hydrogen gas to produce ammonia gas, in the presence of a catalyst. If the equilibrium constant for this reaction at a certain temperature is 6.2 × 10^-2 and the initial concentrations of nitrogen and hydrogen gases are 1.2 M and 0.8 M respectively, what would be the equilibrium concentrations of the reactants and products if the catalyst increases the rate of the forward reaction by a factor of 4?

Question

An industrial process involves the reaction of nitrogen gas and hydrogen gas to produce ammonia gas, in the presence of a catalyst. If the equilibrium constant for this reaction at a certain temperature is 6.2 × 10^-2 and the initial concentrations of nitrogen and hydrogen gases are 1.2 M and 0.8 M respectively, what would be the equilibrium concentrations of the reactants and products if the catalyst increases the r

Answer 1

To solve this problem, we need to use the equilibrium constant expression for the reaction and the initial concentrations of the reactants. The balanced equation for the reaction is:N g  + 3H g   2NH g The equilibrium constant expression is:Kc = [NH] /  [N] * [H] Given that the initial concentrations of nitrogen and hydrogen gases are 1.2 M and 0.8 M respectively, we can set up an ICE  Initial, Change, Equilibrium  table to find the equilibrium concentrations:          N      +    3H           2NHInitial:  1.2 M         0.8 M            0 MChange:  -x M        -3x M          +2x MEquilibrium:  1.2-x  M  0.8-3x  M   2x MNow, we can plug these equilibrium concentrations into the equilibrium constant expression:6.2  10^-2 =  2x  /   1.2-x  *  0.8-3x  However, it is important to note that the catalyst increases the rate of the forward reaction by a factor of 4. This means that the reaction will reach equilibrium faster, but it does not affect the equilibrium concentrations themselves. The equilibrium constant remains the same, as it is only affected by temperature changes.So, we can solve the equation for x:6.2  10^-2 =  2x  /   1.2-x  *  0.8-3x  This equation is quite complex to solve directly, so we can use numerical methods or specialized software to find the value of x. After solving for x, we get:x  0.054Now we can find the equilibrium concentrations:[N] = 1.2 - x  1.2 - 0.054  1.146 M[H] = 0.8 - 3x  0.8 - 3 0.054   0.638 M[NH] = 2x  2 0.054   0.108 MSo, the equilibrium concentrations are approximately 1.146 M for N, 0.638 M for H, and 0.108 M for NH.