An amount of 3 moles of and some is introduced into an evacuated vessel. The reaction starts at t = — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - State the Law of Conservation of Mass According to the **Law of Conservation of Mass**, the total mass of a closed chemical system remains constant at all times during a chemical reaction. Therefore, for the synthesis of ammonia: \[\ce{N2(g) + 3H2(g) <=> 2NH3(g)}\] The sum of the masses of all components inside the vessel must remain constant at any time \(t\): \[w_{\text{total}} = w(\ce{N2}) + w(\ce{H2}) + w(\ce{NH3}) = \text{constant}\] Step 2 - Determine the initial mass of Nitrogen (\(\ce{N2}\)) \[w(\ce{N2})_{\text{initial}} = 3\text{ mol} \times 28\text{ g/mol} = 84\text{ g}\] Step 3 - Determine the initial mass of Hydrogen (\(\ce{H2}\)) Let initial moles of \(\ce{H2}\) = \(a\). Let \(y\) moles of \(\ce{N2}\) react at time \(t\): \[w(\ce{N2}) = 28(3 - y)\text{ g}, \quad w(\ce{H2}) = 2(a - 3y)\text{ g}\] The mass ratio is constant at \(\frac{14}{3}\): \[\frac{w(\ce{N2})}{w(\ce{H2})} = \frac{28(3 - y)}{2(a - 3y)} = \frac{14(3-y)}{a - 3y} = \frac{14}{3}\] For this to be independent of \(y\): \[\frac{3 - y}{a - 3y} = \frac{1}{3} \implies 3(3-y) = a - 3y \implies 9 - 3y = a - 3y \implies a = 9\text{ mol}\] \[w(\ce{H2})_{\text{initial}} = 9\text{ mol} \times 2\text{ g/mol} = 18\text{ g}\] Step 4 - Calculate the total mass of the system \[w_{\text{total}} = 84\text{ g} + 18\text{ g} + 0\text{ g} = \boxed{102\text{ g}}\] This total mass is conserved at all times. Step 5 - Evaluate all Options - **Option (A) 118 g at \(t = t_1\)**: Incorrect. Total mass is always 102 g, not 118 g. - **Option (B) 102 g at \(t = 2t_1\)**: Correct. Mass is conserved at 102 g at all times, including \(t = 2t_1\). - **Option (C) 50 g at \(t = t_1/3\)**: Incorrect. Total mass at any time must be 102 g. - **Option (D) cannot be predicted**: Incorrect. The Law of Conservation of Mass guarantees the total mass is always 102 g.