Question

3. Consider the combustion of methane at 298 K, $$CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)$$ $$\Delta H = -890.4 kJ$$ Which one of the following best explains why this reaction is spontaneous at 298 K? A) $$\Delta S_{sys} > 0; \Delta S_{surr} < 0$$ B) $$\Delta S_{sys} < 0; \Delta S_{surr} < 0$$ C) $$\Delta S_{sys} \approx 0; \Delta S_{surr} > 0$$ D) $$\Delta S_{sys} > 0; \Delta S_{surr} \approx 0$$ E) $$\Delta S_{sys} \approx 0; \Delta S_{surr} \approx 0$$

Name: 3 consider the combustion of methane at 298 k ch4g 2o2g rightarrow co2g 2h2og delta h 8904 kj which one of the following best explains why this reaction is spontaneous at 298 k a delta s 10046
Uploaded: 2023-06-07T15:42:03-08:00
Duration: 2 min 21 s
Channel: Madhur L
Description: 3 consider the combustion of methane at 298 k ch4g 2o2g rightarrow co2g 2h2og delta h 8904 kj which one of the following best explains why this reaction is spontaneous at 298 k a delta s 10046

          3. Consider the combustion of methane at 298 K,
$$CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)$$
$$\Delta H = -890.4 kJ$$
Which one of the following best explains why this reaction is spontaneous at 298 K?
A) $$\Delta S_{sys} > 0; \Delta S_{surr} < 0$$
B) $$\Delta S_{sys} < 0; \Delta S_{surr} < 0$$
C) $$\Delta S_{sys} \approx 0; \Delta S_{surr} > 0$$
D) $$\Delta S_{sys} > 0; \Delta S_{surr} \approx 0$$
E) $$\Delta S_{sys} \approx 0; \Delta S_{surr} \approx 0$$

3 consider the combustion of methane at 298 k ch4g 2o2g rightarrow co2g 2h2og delta h 8904 kj which one of the following best explains why this reaction is spontaneous at 298 k a delta s 10046

Added by Monica O.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Madhur L

06/07/2023

Step 1

We know that $$\Delta G = \Delta H - T\Delta S$$. Since $$\Delta H = -890.4 kJ < 0$$, the reaction is exothermic. Show more…

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3. Consider the combustion of methane at 298 K, $$CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)$$ $$\Delta H = -890.4 kJ$$ Which one of the following best explains why this reaction is spontaneous at 298 K? A) $$\Delta S_{sys} > 0; \Delta S_{surr} < 0$$ B) $$\Delta S_{sys} < 0; \Delta S_{surr} < 0$$ C) $$\Delta S_{sys} \approx 0; \Delta S_{surr} > 0$$ D) $$\Delta S_{sys} > 0; \Delta S_{surr} \approx 0$$ E) $$\Delta S_{sys} \approx 0; \Delta S_{surr} \approx 0$$