The graph below shows concentration changes of reactants and products for this reaction over time. Use this information to answer the following questions:

Question

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 · 2026-06-01T11:02:42.171Z

Here are the answers to your questions: 1.1 Explain the term chemical equilibrium. Chemical equilibrium is a state in a reversible reaction where the rate of the forward reaction is equal to the rate of the reverse reaction, and the concentrations of reactants and products remain constant. 1.2 At which times/interval(s) is / (are) the reaction at equilibrium? The reaction is at equilibrium during the intervals 0 to t_1, t_1 to t_2, and t_3 to t_4. 1.3 Does the FORWARD or the REVERSE reaction have the faster reaction rate during the interval t_2 to t_3? During the interval t_2 to t_3, the concentration of reactants ([CO] and [O_2]) decreases, and the concentration of the product ([CO_2]) increases. This indicates that the forward reaction rate is faster. 1.4 The temperature is decreased at t_4. Is the reaction EXOTHERMIC or ENDOTHERMIC? At t_4, the temperature is decreased. The graph shows that after this change, the concentrations of CO and O_2 decrease, while the concentration of CO_2 increases. This means the equilibrium shifts to the right (forward reaction). According to Le Chatelier's principle, a decrease in temperature favors the exothermic reaction. Therefore, the forward reaction is exothermic. 1.5 Explain the answer to QUESTION 1.4. When the temperature is decreased at t_4, the system shifts to counteract this change by favoring the reaction that releases heat. Since the equilibrium shifts to the right, increasing the concentration of CO_2 and decreasing the concentrations of CO and O_2, the forward reaction must be exothermic, as it releases heat. 1.6 What change is made to at t_2? At t_2, there is an instantaneous increase in the concentrations of all reactants and products. This indicates a decrease in the volume of the container (or an increase in pressure). 1.7 How will the change made at t_2 influence the value of K_c for this reaction? Write only INCREASE, DECREASE or REMAIN THE SAME. The change made at t_2 (a change in pressure/volume) does not affect the value of K_c. K_c is only influenced by temperature. Therefore, K_c will remain the same. 1.8 Initially 63 g of CO and 9.11 g of O_2 is placed in a 2 dm^3 container. At equilibrium, the concentration of CO_2 is 0.15 mol·dm^-3. Calculate the K_c value for this reaction. Step 1: Calculate the initial moles and concentrations of reactants. Molar mass of CO (M_CO) = 12.01 + 16.00 = 28.01 g/mol Molar mass of O_2 (M_O_2) = 2 × 16.00 = 32.00 g/mol Initial moles of CO (n_CO) = 63 g28.01 g/mol = 2.2492 mol Initial moles of O_2 (n_O_2) = 9.11 g32.00 g/mol = 0.2847 mol Volume of container (V) = 2 dm^3 Initial concentration of CO ([CO]_initial) = 2.2492 mol2 dm^3 = 1.1246 mol/dm^3 Initial concentration of O_2 ([O_2]_initial) = 0.2847 mol2 dm^3 = 0.1424 mol/dm^3 Initial concentration of CO_2 ([CO_2]_initial) = 0 mol/dm^3 Step 2: Set up an ICE (Initial, Change, Equilibrium) table using concentrations. The balanced reaction is: 2CO(g) + O_2(g) 2CO_2(g) | Species | Initial (mol/dm^3) | Change (mol/dm^3) | Equilibrium (mol/dm^3) | | :------ | :------------------- | :------------------ | :----------------------- | | CO | 1.1246 | -2x | 1.1246 - 2x | | O_2 | 0.1424 | -x | 0.1424 - 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. From the table, 2x = 0.15 mol/dm^3. So, x = (0.15)/(2) = 0.075 mol/dm^3. Step 4: Calculate the equilibrium concentrations of all species. [CO]_eq = 1.1246 - 2(0.075) = 1.1246 - 0.15 = 0.9746 mol/dm^3 [O_2]_eq = 0.1424 - 0.075 = 0.0674 mol/dm^3 [CO_2]_eq = 0.15 mol/dm^3 Step 5: Calculate the equilibrium constant K_c. The expression for K_c is: K_c = ([CO_2]^2)/([CO]^2[O_2]) Substitute the equilibrium concentrations: K_c = ((0.15)^2)/((0.9746)^2(0.0674)) K_c = (0.0225)/((0.9498)(0.0674)) K_c = (0.0225)/(0.064026) K_c ≈ 0.3514 Rounding to three significant figures: K_c = 0.351 That's 2 down. 3 left today — send the next one.

Accepted Answer · 2026-06-01T11:02:42.171Z

Here are the answers to your questions: 1.1 Explain the term chemical equilibrium. Chemical equilibrium is a state in a reversible reaction where the rate of the forward reaction is equal to the rate of the reverse reaction, and the concentrations of reactants and products remain constant. 1.2 At which times/interval(s) is / (are) the reaction at equilibrium? The reaction is at equilibrium during the intervals 0 to t_1, t_1 to t_2, and t_3 to t_4. 1.3 Does the FORWARD or the REVERSE reaction have the faster reaction rate during the interval t_2 to t_3? During the interval t_2 to t_3, the concentration of reactants ([CO] and [O_2]) decreases, and the concentration of the product ([CO_2]) increases. This indicates that the forward reaction rate is faster. 1.4 The temperature is decreased at t_4. Is the reaction EXOTHERMIC or ENDOTHERMIC? At t_4, the temperature is decreased. The graph shows that after this change, the concentrations of CO and O_2 decrease, while the concentration of CO_2 increases. This means the equilibrium shifts to the right (forward reaction). According to Le Chatelier's principle, a decrease in temperature favors the exothermic reaction. Therefore, the forward reaction is exothermic. 1.5 Explain the answer to QUESTION 1.4. When the temperature is decreased at t_4, the system shifts to counteract this change by favoring the reaction that releases heat. Since the equilibrium shifts to the right, increasing the concentration of CO_2 and decreasing the concentrations of CO and O_2, the forward reaction must be exothermic, as it releases heat. 1.6 What change is made to at t_2? At t_2, there is an instantaneous increase in the concentrations of all reactants and products. This indicates a decrease in the volume of the container (or an increase in pressure). 1.7 How will the change made at t_2 influence the value of K_c for this reaction? Write only INCREASE, DECREASE or REMAIN THE SAME. The change made at t_2 (a change in pressure/volume) does not affect the value of K_c. K_c is only influenced by temperature. Therefore, K_c will remain the same. 1.8 Initially 63 g of CO and 9.11 g of O_2 is placed in a 2 dm^3 container. At equilibrium, the concentration of CO_2 is 0.15 mol·dm^-3. Calculate the K_c value for this reaction. Step 1: Calculate the initial moles and concentrations of reactants. Molar mass of CO (M_CO) = 12.01 + 16.00 = 28.01 g/mol Molar mass of O_2 (M_O_2) = 2 × 16.00 = 32.00 g/mol Initial moles of CO (n_CO) = 63 g28.01 g/mol = 2.2492 mol Initial moles of O_2 (n_O_2) = 9.11 g32.00 g/mol = 0.2847 mol Volume of container (V) = 2 dm^3 Initial concentration of CO ([CO]_initial) = 2.2492 mol2 dm^3 = 1.1246 mol/dm^3 Initial concentration of O_2 ([O_2]_initial) = 0.2847 mol2 dm^3 = 0.1424 mol/dm^3 Initial concentration of CO_2 ([CO_2]_initial) = 0 mol/dm^3 Step 2: Set up an ICE (Initial, Change, Equilibrium) table using concentrations. The balanced reaction is: 2CO(g) + O_2(g) 2CO_2(g) | Species | Initial (mol/dm^3) | Change (mol/dm^3) | Equilibrium (mol/dm^3) | | :------ | :------------------- | :------------------ | :----------------------- | | CO | 1.1246 | -2x | 1.1246 - 2x | | O_2 | 0.1424 | -x | 0.1424 - 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. From the table, 2x = 0.15 mol/dm^3. So, x = (0.15)/(2) = 0.075 mol/dm^3. Step 4: Calculate the equilibrium concentrations of all species. [CO]_eq = 1.1246 - 2(0.075) = 1.1246 - 0.15 = 0.9746 mol/dm^3 [O_2]_eq = 0.1424 - 0.075 = 0.0674 mol/dm^3 [CO_2]_eq = 0.15 mol/dm^3 Step 5: Calculate the equilibrium constant K_c. The expression for K_c is: K_c = ([CO_2]^2)/([CO]^2[O_2]) Substitute the equilibrium concentrations: K_c = ((0.15)^2)/((0.9746)^2(0.0674)) K_c = (0.0225)/((0.9498)(0.0674)) K_c = (0.0225)/(0.064026) K_c ≈ 0.3514 Rounding to three significant figures: K_c = 0.351 That's 2 down. 3 left today — send the next one.

The graph below shows concentration changes of reactants and products for this reaction over time. Use this information to answer the following questions:

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The graph below shows concentration changes of reactants and products for this reaction over time. Use this information to answer the following questions:

Need help with your own homework?

More Chemistry Questions