QuestionΔΗ -1237. kJ J AS = 2987. K 6C(s) + 6H2(g) + 302(g) → C6H12O6 (s) AG = kJ Which is spontaneous? this reaction the reverse reaction 4H3PO4(s) P4O10 (s) + 6H₂O(1) Oneither ΔΗ
439. kJ J AS K AG= -34. kJ Which is spontaneous? this reaction O the reverse reaction Oneither

Studdy Solution

STEP 1

What is this asking? We need to figure out if two chemical reactions happen spontaneously, and if so, in which direction. Watch out! Don't forget to convert units and watch those signs!
A negative $\Delta G$ means spontaneous, a positive $\Delta G$ means the reverse is spontaneous, and a $\Delta G$ of zero means neither!

STEP 2

1. Calculate Gibbs Free Energy for the first reaction.
2. Determine spontaneity for the first reaction.
3. Determine spontaneity for the second reaction.

STEP 3

First, we need to make sure our units match up!
We've got $\Delta S$ in Joules per Kelvin, but $\Delta H$ is in Kilojoules.
Let's convert $\Delta S$ to Kilojoules per Kelvin.
We can do this by dividing by 1000, since there are 1000 Joules in a Kilojoule.
So, $\Delta S = \frac{2987 \text{ J/K}}{1000 \text{ J/kJ}} = 2.987 \text{ kJ/K}$ .

STEP 4

Now, we can use the Gibbs Free Energy equation: $\Delta G = \Delta H - T\Delta S$ We aren't given a temperature, so we'll assume standard temperature, which is 298 K.
Let's plug in our values: $\Delta G = -1237 \text{ kJ} - (298 \text{ K} \cdot 2.987 \text{ kJ/K})$

STEP 5

$\Delta G = -1237 \text{ kJ} - 890.126 \text{ kJ}$ $\Delta G = -2127.126 \text{ kJ}$ So, $\Delta G$ for the first reaction is approximately -2127 kJ.

STEP 6

Since $\Delta G$ is negative, the reaction is spontaneous as written!

STEP 7

We're given that $\Delta G$ for the second reaction is -34 kJ.
Since this value is negative, this reaction is also spontaneous as written!

STEP 8

The first reaction ( $6\text{C(s)} + 6\text{H}_2\text{(g)} + 3\text{O}_2\text{(g)} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6\text{(s)}$ ) is spontaneous.
The second reaction ( $4\text{H}_3\text{PO}_4\text{(s)} \rightarrow \text{P}_4\text{O}_{10}\text{(s)} + 6\text{H}_2\text{O(l)}$ ) is also spontaneous!

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QuestionΔΗ -1237. kJ J AS = 2987. K 6C(s) + 6H2(g) + 302(g) → C6H12O6 (s) AG = kJ Which is spontaneous? this reaction the reverse reaction 4H3PO4(s) P4O10 (s) + 6H₂O(1) Oneither ΔΗ
439. kJ J AS K AG= -34. kJ Which is spontaneous? this reaction O the reverse reaction Oneither

Studdy Solution

STEP 1

STEP 2

1. Calculate Gibbs Free Energy for the first reaction.
2. Determine spontaneity for the first reaction.
3. Determine spontaneity for the second reaction.

STEP 3

STEP 4

Now, we can use the Gibbs Free Energy equation: $\Delta G = \Delta H - T\Delta S$ We aren't given a temperature, so we'll assume standard temperature, which is 298 K.
Let's plug in our values: $\Delta G = -1237 \text{ kJ} - (298 \text{ K} \cdot 2.987 \text{ kJ/K})$

STEP 5

$\Delta G = -1237 \text{ kJ} - 890.126 \text{ kJ}$ $\Delta G = -2127.126 \text{ kJ}$ So, $\Delta G$ for the first reaction is approximately -2127 kJ.

STEP 6

Since $\Delta G$ is negative, the reaction is spontaneous as written!

STEP 7

We're given that $\Delta G$ for the second reaction is -34 kJ.
Since this value is negative, this reaction is also spontaneous as written!

STEP 8

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QuestionΔΗ -1237. kJ J AS = 2987. K 6C(s) + 6H2(g) + 302(g) → C6H12O6 (s) AG = kJ Which is spontaneous? this reaction the reverse reaction 4H3PO4(s) P4O10 (s) + 6H₂O(1) Oneither ΔΗ439. kJ J AS K AG= -34. kJ Which is spontaneous? this reaction O the reverse reaction Oneither

Studdy Solution

STEP 1

STEP 2

1. Calculate Gibbs Free Energy for the first reaction.2. Determine spontaneity for the first reaction.3. Determine spontaneity for the second reaction.

STEP 3

STEP 4

STEP 5

ΔG=−1237 kJ−890.126 kJ \Delta G = -1237 \text{ kJ} - 890.126 \text{ kJ} ΔG=−1237 kJ−890.126 kJ ΔG=−2127.126 kJ \Delta G = -2127.126 \text{ kJ} ΔG=−2127.126 kJ So, ΔG\Delta GΔG for the first reaction is approximately **-2127 kJ**.

STEP 6

Since ΔG\Delta GΔG is **negative**, the reaction is spontaneous as written!

STEP 7

We're given that ΔG\Delta GΔG for the second reaction is **-34 kJ**.Since this value is **negative**, this reaction is also spontaneous as written!

STEP 8

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QuestionΔΗ -1237. kJ J AS = 2987. K 6C(s) + 6H2(g) + 302(g) → C6H12O6 (s) AG = kJ Which is spontaneous? this reaction the reverse reaction 4H3PO4(s) P4O10 (s) + 6H₂O(1) Oneither ΔΗ
439. kJ J AS K AG= -34. kJ Which is spontaneous? this reaction O the reverse reaction Oneither

1. Calculate Gibbs Free Energy for the first reaction.
2. Determine spontaneity for the first reaction.
3. Determine spontaneity for the second reaction.

$\Delta G = -1237 \text{ kJ} - 890.126 \text{ kJ}$ $\Delta G = -2127.126 \text{ kJ}$ So, $\Delta G$ for the first reaction is approximately -2127 kJ.

Since $\Delta G$ is negative, the reaction is spontaneous as written!

We're given that $\Delta G$ for the second reaction is -34 kJ.
Since this value is negative, this reaction is also spontaneous as written!