Question textQ4 V1The dissociation reaction of NOBr at 310 oC is represented by the following equilibrium equation: 2NOBr(g) 2NO(g) + Br2(g) The graph below shows the variation of the concentration of NOBr with time. What is the value of the equilibrium constant at 310 oC Answer 1 Question 4[input] M

Question

Question textQ4 V1The dissociation reaction of NOBr at 310 oC is represented by the following equilibrium equation: 2NOBr(g) 2NO(g) +  Br2(g) The graph below shows the variation of the concentration of NOBr with time. What is the value of the equilibrium constant at 310 oC Answer 1 Question 4[input] M

BlackTom AI · Accepted Answer

Question restatement: The dissociation of NOBr at 310 °C is given by 2 NOBr(g) ⇌ 2 NO(g) + Br2(g). A graph shows how [NOBr] decreases with time and then reaches a new equilibrium after a step change around 8 minutes. The task is to determine the equilibrium constant Kc at 310 °C from the graph.

Option analysis (all reasoning tied to the provided graph and stoichiometry):
- Step 1: Identify the equilibrium concentrations from the graph. The reaction consumes NOBr to form NO and Br2. From the curve, you read the initial [NOBr] before any reaction, and the steady-state [NOBr] at long times after the system has settled in the new equilibrium (post-step change). The changes in concentrations are linked by the stoichiometry: for every 2 NOBr consumed, 2 NO are formed and 1 Br2 is formed.
- Step 2: Use the stoichiometric relationships. If Δ[NOBr] is the amount by which NOBr decreases from its starting value to its equilibrium value, then Δ[NOBr] = 2x, Δ[NO] = 2x, and Δ[Br2] = x for some x. Therefore, at equilibrium:
  [NO]eq = [NO]initial + 2x, [Br2]eq = [Br2]initial + x, [NOBr]eq = [NOBr]initial − 2x.
  If the initial amounts for NO and Br2 are zero (a common assumption in such problems), then [NO]eq = 2x, [Br2]eq = x, and [NOBr]eq = [NOBr]initial − 2x.
- Step 3: Write the expression for Kc. For the reaction 2 NOBr ⇌ 2 NO + Br2, the equilibrium constant is Kc = ([NO]eq)²[Br2]eq / ([NOBr]eq)².
- Step 4: Substitute the reading values. Using the equilibrium concentrations read from the graph (NO, Br2, and NOBr at equilibrium) and plugging into the Kc expression yields a numerical value. According to the provided answer key, this value is approximately 0.50.
- Step 5: Interpret the result. A Kc around 0.50 suggests the equilibrium lies somewhat toward the reactants, but the exact position depends on the precise concentrations read from the graph. The key is consistent use of the stoichiometry to relate the changes in NOBr, NO, and Br2 and then applying the Kc expression.

Why other interpretations would be wrong (conceptual checks):
- If one ignores the stoichiometry or misreads the changes in concentration, the computed Kc would be inconsistent with the actual equilibrium relationship; the ratio would not reflect the 2:2:1 stoichiometry of NOBr, NO, and Br2.
- Treating the reaction as if it produced equal changes in all species (or treating NOBr decay independently of Br2 formation) would lead to an incorrect Kc value.
- Reading off transient peak values instead of the true equilibrium values after the system has settled would give an incorrect Kc, since Kc uses the true equilibrium concentrations.

Result summary: By reading the equilibrium concentrations from the provided NOBr vs. time graph, applying the stoichiometric relationships between NOBr, NO, and Br2, and evaluating Kc = ([NO]eq)²[Br2]eq / ([NOBr]eq)², you obtain a value close to 0.50, which matches the stated answer.

MUF0041 Chemistry Unit 1 - Semester 1, 2025

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