Hydrogen is produced commercially by the reaction of methane...

Hydrogen is produced commercially by the reaction of methane with steam: $$\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(g)$$ a. Calculate $\Delta H^{\circ}$ and $\Delta S^{\circ}$ for this reaction (use the data in Appendix 4 ). b. What temperatures will favor product formation assuming standard conditions and assuming that $\Delta H^{\circ}$ and $\Delta S^{\circ}$ do not depend on temperature?

Hydrogen is produced commercially by the reaction of methane with steam:
$$\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(g)$$
a. Calculate $\Delta H^{\circ}$ and $\Delta S^{\circ}$ for this reaction (use the data in Appendix 4 ).
b. What temperatures will favor product formation assuming standard conditions and assuming that $\Delta H^{\circ}$ and $\Delta S^{\circ}$ do not depend on temperature?

Key Concepts

Standard Enthalpy Change (?H°)

This concept involves calculating the overall enthalpy change of a chemical reaction using standard enthalpy of formation values for the reactants and products. The process entails subtracting the sum of the enthalpies of the reactants from the sum of the enthalpies of the products, reflecting the energy absorbed or released during the reaction at standard conditions.

Standard Entropy Change (?S°)

Standard entropy change is determined by calculating the difference between the total entropy of the products and that of the reactants, using their standard molar entropy values. This measure helps indicate the degree of disorder or randomness change that accompanies a chemical reaction under standard conditions.

Gibbs Free Energy (?G)

Gibbs free energy combines the effects of enthalpy and entropy to predict reaction spontaneity, following the equation ?G = ?H – T?S. A negative ?G under specified conditions implies that the reaction is spontaneous, making this concept fundamental for assessing whether a chemical reaction will favor the formation of products or reactants.

Temperature Dependence of Reaction Equilibrium

The influence of temperature on reaction equilibrium is critical, especially when considering the signs of ?H and ?S. Temperature changes shift the balance between the energetic and entropic contributions in the free energy equation, which in turn can favor either the forward or the reverse reaction. For example, in reactions where a positive ?S supports product formation, increasing temperature can drive the reaction toward products.

Transcript

00:01 Okay, so we have the following balance reaction, and for part a, they want us to calculate the delta h and the delta s.

00:09 So let's start with the delta h.

00:12 So the delta h of a reaction is equal to the sum of the delta h on the product side, minus the sum of the delta h on the react inside.

00:27 Right, so let's start calculating this.

00:29 So delta h of the reaction, so we let's start with the product side.

00:34 So we have co on the product side.

00:37 We have one mole of co.

00:40 Let's look at the delta h value.

00:41 So in this textbook, the delta h value is located in appendix 4.

00:46 So we see the co has a delta h of negative 110 .5 kilojoules.

00:54 Now let's look at h2.

00:57 Oh, h2, the state is gas.

01:01 So we have three modes of h2 and the delta h is zero kilojoules.

01:06 Subtract, now let's look at the reacting size.

01:09 So we have one mole of ch4.

01:12 Ch4 has a delta h of negative 75 kilojoules plus delta h, no, the delta h2o.

01:24 So we have one mole of h2o times it by the negative 242.

01:30 Okay, so let's plug all this, once we plugged all this into a calculator, we should get delta h of the reaction is equal to 206.

01:40 0 .5 kilojoules.

01:42 So it's an endodermic reaction.

01:45 So next part they want us to calculate delta s right so for a delta s we pretty much do the same thing we take the product and we subtract the react from it, right? so so delta s you go to the sum of the delta s on the product side minus the sum of the delta s on the react inside.

02:23 Okay, so let's start.

02:25 So delta s of the reaction is equal to, so on the product side, we have co, 1 mole of co times it by 198, plus 3 moles of h2, right? so 3 times 131, minus 1 moa of ch4, so 186, plus 1 mole of h2l, so times it by 189...

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