Explain the term chemical equilibrium. At which times interval(s) is / (are) the reaction at equilibrium? Does the FORWARD or the REVERSE reaction have the faster reaction rate during the interval t2 to t3?

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Asked on June 1, 2026Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

Accepted Answer · 2026-06-01T10:57:53.800Z

Here are the solutions to your questions: 1.1 Chemical equilibrium is a state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant. 1.2 The reaction is at equilibrium during the time intervals where the concentrations of all reactants and products are constant. The concentrations are constant during the intervals: t_1 to t_2 and t_3 to t_4 1.3 During the interval t_2 to t_3, the concentration of CO_2 (product) decreases, and the concentrations of CO and O_2 (reactants) increase. This indicates that the reverse reaction is favored. REVERSE reaction 1.4 At t_4, the temperature is decreased. The graph shows that after t_4, the concentration of CO_2 increases, and the concentrations of CO and O_2 decrease. This means the forward reaction is favored. Since a decrease in temperature favors the forward reaction, the forward reaction must be exothermic. EXOTHERMIC 1.5 When the temperature is decreased, the system shifts to favor the reaction that releases heat (exothermic reaction) to counteract the change. Since the forward reaction is favored (indicated by the increase in product CO_2 and decrease in reactants CO and O_2), the forward reaction must be exothermic. 1.6 At t_2, the concentration of CO_2 (product) decreases sharply, while the concentrations of CO and O_2 (reactants) increase sharply. This indicates that some of the product, CO_2, was removed from the system. Removal of CO_2 1.7 The value of K_c for a given reaction is only affected by temperature. Since the change made at t_2 (removal of CO_2) is not a temperature change, the value of K_c will remain the same. REMAIN THE SAME 1.8 Step 1: Calculate the initial moles of CO and O_2. Molar mass of CO = 12.01 + 16.00 = 28.01 g/mol Molar mass of O_2 = 2 × 16.00 = 32.00 g/mol Initial moles of CO = 63 g28.01 g/mol = 2.249 mol Initial moles of O_2 = 9.11 g32.00 g/mol = 0.2847 mol Step 2: Set up an ICE table using initial concentrations. Volume of container = 2 dm^3 Initial concentration of CO = 2.249 mol2 dm^3 = 1.1245 mol/dm^3 Initial concentration of O_2 = 0.2847 mol2 dm^3 = 0.14235 mol/dm^3 Initial concentration of CO_2 = 0 mol/dm^3 The balanced equation is: 2CO(g) + O_2(g) 2CO_2(g) | Species | Initial (mol/dm^3) | Change (mol/dm^3) | Equilibrium (mol/dm^3) | | :------ | :-------------------- | :------------------- | :------------------------ | | CO | 1.1245 | -2x | 1.1245 - 2x | | O_2 | 0.14235 | -x | 0.14235 - x | | CO_2 | 0 | +2x | 2x | Step 3: Use the given equilibrium concentration of CO_2 to find x. At equilibrium, [CO_2] = 0.15 mol/dm^3. So, 2x = 0.15 mol/dm^3 x = (0.15)/(2) = 0.075 mol/dm^3 Step 4: Calculate the equilibrium concentrations of CO and O_2. [CO] = 1.1245 - 2x = 1.1245 - 2(0.075) = 1.1245 - 0.15 = 0.9745 mol/dm^3 [O_2] = 0.14235 - x = 0.14235 - 0.075 = 0.06735 mol/dm^3 Step 5: Write the expression for K_c and substitute the equilibrium concentrations. K_c = [CO_2]^2[CO]^2[O_2] K_c = ((0.15)^2)/((0.9745)^2(0.06735)) K_c = (0.0225)/((0.94965)(0.06735)) K_c = (0.0225)/(0.06402) K_c = 0.35145 Rounding to three significant figures: K_c = 0.351 Send me the next one 📸

Accepted Answer · 2026-06-01T10:57:53.800Z

Here are the solutions to your questions: 1.1 Chemical equilibrium is a state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant. 1.2 The reaction is at equilibrium during the time intervals where the concentrations of all reactants and products are constant. The concentrations are constant during the intervals: t_1 to t_2 and t_3 to t_4 1.3 During the interval t_2 to t_3, the concentration of CO_2 (product) decreases, and the concentrations of CO and O_2 (reactants) increase. This indicates that the reverse reaction is favored. REVERSE reaction 1.4 At t_4, the temperature is decreased. The graph shows that after t_4, the concentration of CO_2 increases, and the concentrations of CO and O_2 decrease. This means the forward reaction is favored. Since a decrease in temperature favors the forward reaction, the forward reaction must be exothermic. EXOTHERMIC 1.5 When the temperature is decreased, the system shifts to favor the reaction that releases heat (exothermic reaction) to counteract the change. Since the forward reaction is favored (indicated by the increase in product CO_2 and decrease in reactants CO and O_2), the forward reaction must be exothermic. 1.6 At t_2, the concentration of CO_2 (product) decreases sharply, while the concentrations of CO and O_2 (reactants) increase sharply. This indicates that some of the product, CO_2, was removed from the system. Removal of CO_2 1.7 The value of K_c for a given reaction is only affected by temperature. Since the change made at t_2 (removal of CO_2) is not a temperature change, the value of K_c will remain the same. REMAIN THE SAME 1.8 Step 1: Calculate the initial moles of CO and O_2. Molar mass of CO = 12.01 + 16.00 = 28.01 g/mol Molar mass of O_2 = 2 × 16.00 = 32.00 g/mol Initial moles of CO = 63 g28.01 g/mol = 2.249 mol Initial moles of O_2 = 9.11 g32.00 g/mol = 0.2847 mol Step 2: Set up an ICE table using initial concentrations. Volume of container = 2 dm^3 Initial concentration of CO = 2.249 mol2 dm^3 = 1.1245 mol/dm^3 Initial concentration of O_2 = 0.2847 mol2 dm^3 = 0.14235 mol/dm^3 Initial concentration of CO_2 = 0 mol/dm^3 The balanced equation is: 2CO(g) + O_2(g) 2CO_2(g) | Species | Initial (mol/dm^3) | Change (mol/dm^3) | Equilibrium (mol/dm^3) | | :------ | :-------------------- | :------------------- | :------------------------ | | CO | 1.1245 | -2x | 1.1245 - 2x | | O_2 | 0.14235 | -x | 0.14235 - x | | CO_2 | 0 | +2x | 2x | Step 3: Use the given equilibrium concentration of CO_2 to find x. At equilibrium, [CO_2] = 0.15 mol/dm^3. So, 2x = 0.15 mol/dm^3 x = (0.15)/(2) = 0.075 mol/dm^3 Step 4: Calculate the equilibrium concentrations of CO and O_2. [CO] = 1.1245 - 2x = 1.1245 - 2(0.075) = 1.1245 - 0.15 = 0.9745 mol/dm^3 [O_2] = 0.14235 - x = 0.14235 - 0.075 = 0.06735 mol/dm^3 Step 5: Write the expression for K_c and substitute the equilibrium concentrations. K_c = [CO_2]^2[CO]^2[O_2] K_c = ((0.15)^2)/((0.9745)^2(0.06735)) K_c = (0.0225)/((0.94965)(0.06735)) K_c = (0.0225)/(0.06402) K_c = 0.35145 Rounding to three significant figures: K_c = 0.351 Send me the next one 📸

Explain the term chemical equilibrium. At which times interval(s) is / (are) the reaction at equilibrium? Does the FORWARD or the REVERSE reaction have the faster reaction rate during the interval t2 to t3?

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Explain the term chemical equilibrium. At which times interval(s) is / (are) the reaction at equilibrium? Does the FORWARD or the REVERSE reaction have the faster reaction rate during the interval t2 to t3?

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