Question

Use Hess's Law to calculate the enthalpy change for recovering tungsten from its oxide using the...

Use Hess's Law to calculate the enthalpy change for recovering tungsten from its oxide using the reaction:WO3(s) + 3H2(g) --> W(s)+3H2O(g)

from the following data :

2W(s)+ 3O2(g)-->2WO3(s) , change in H = -1685.4 kJ

2H2(g)+O2(g)---> 2H2O(g) , change in H = -477.84 kJ

Science   Chemistry   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

Target reaction :   WO3(s) + 3H2(g) --------> W(s)+3H2O(g)

Given reactions :

2W(s)+ 3O2(g)-->2WO3(s) , change in H = -1685.4 kJ      Reaction1

2H2(g)+O2(g)---> 2H2O(g) , change in H = -477.84 kJ    reaction2

so carrying out the algebraic operations on the given reactions to get the target reaction.

-1/2(Reaction1 ) + 3/2( reaction2 ) = target reaction

so Enthapy of target reaction = -1/2( -1685.4 kJ) + 3/2(-477.84 kJ )

                                             = 842.7 + ( -716.76 ) = 125.94 KJ

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Use Hess's Law to calculate the enthalpy change for the reaction: GdO3 + 3H2(g) -> Gd(s)...
Use Hess's Law to calculate the enthalpy change for the reaction: GdO3 + 3H2(g) -> Gd(s) + H2O (g) from the following data: 2Gd(s) + 3O2 --> 2GdO3(s)   ^H=-685.4kJ 2H2 (g) + O2 (g) --> 2H2O(g)     ^H= -47.8kJ
We can use Hess's law to calculate enthalpy changes that cannot be measured. One such reaction...
We can use Hess's law to calculate enthalpy changes that cannot be measured. One such reaction is the conversion of methane to ethylene: 2CH4(g)⟶C2H4(g)+2H2(g) Calculate the ΔH∘ for this reaction using the following thermochemical data: CH4(g)+2O2(g)⟶CO2(g)+2H2O(l) ΔH∘=−890.3kJ C2H4(g)+H2(g)⟶C2H6(g) ΔH∘=−136.3kJ 2H2(g)+O2(g)⟶2H2O(l) ΔH∘=−571.6kJ 2C2H6(g)+7O2(g)⟶4CO2(g)+6H2O(l ΔH∘=−3120.8kJ
We can use Hess's law to calculate enthalpy changes that cannot be measured. One such reaction...
We can use Hess's law to calculate enthalpy changes that cannot be measured. One such reaction is the conversion of methane to ethylene: 2CH4(g)⟶C2H4(g)+2H2(g) Part A Calculate the ΔH∘ for this reaction using the following thermochemical data: CH4(g)+2O2(g)⟶CO2(g)+2H2O(l) ΔH∘=−890.3kJ C2H4(g)+H2(g)⟶C2H6(g) ΔH∘=−136.3kJ 2H2(g)+O2(g)⟶2H2O(l) ΔH∘=−571.6kJ 2C2H6(g)+7O2(g)⟶4CO2(g)+6H2O(l) ΔH∘=−3120.8kJ Express your answer to four significant figures and include the appropriate units.
Use Hess's Law to calculate the enthalpy of reaction, ΔH rxn, for the reaction in bold...
Use Hess's Law to calculate the enthalpy of reaction, ΔH rxn, for the reaction in bold below given the following chemical steps and their respective enthalpy changes. Show ALL work! 2 C(s) + H2(g) → C2H2(g) ΔH°rxn = ? 1. C2H2(g) + 5/2 O2(g) → 2CO2 (g) + H2O (l) ΔH°rxn = -1299.6 kJ 2. C(s) + O2(g) → CO2 (g) ΔH°rxn = -393.5 kJ 3. H2(g) + ½ O2(g) → H2O (l) ΔH°rxn = -285.8 kJ
Hess's law states that "the heat released or absorbed in a chemical process is the same...
Hess's law states that "the heat released or absorbed in a chemical process is the same whether the process takes place in one or in several steps." It is important to recall the following rules: When two reactions are added, their enthalpy values are added. When a reaction is reversed, the sign of its enthalpy value changes. When the coefficients of a reaction are multiplied by a factor, the enthalpy value is multiplied by that same factor. Part A Calculate...
The standard molar enthalpy of formation, of diborane cannot be determined directly because the compound cannot...
The standard molar enthalpy of formation, of diborane cannot be determined directly because the compound cannot be prepared by reaction of boron and hydrogen. However, the value can be calculated. Calculate the standard enthalpy of formation of gaseous diborane(B2H6) using the following thermochemical information: i) 4B(s) + 3O2(g) --> WB2O3(S) = -2509.1kJ ii) 2H2(g) + O2(g) --> 2H2O(l) = - 571.7kJ iii) B2H6(g) + 3O2(g) --> B2O3(s) + 3H2O(l) = -2147.5 kJ
Using the Information below determine the change in enthalpy for the following reaction: 2NO (g) +...
Using the Information below determine the change in enthalpy for the following reaction: 2NO (g) + 5H2 (g)!2NH3 (g) + 2H2O (l) H2 (g) + 1⁄2O2 (g)!H2O (l) ΔH° = -285.8 kJ N2 (g) + O2 (g)!2NO (l) ΔH° = +180.5 kJ 2NH3 (g)!N2 (g) + 3H2 (g)ΔH° = +92.22 kJ a)-197.52 kJ b)-241.7 kJ c)-483.3 kJ d)-659.88 kJ e)-844.3 kJ please show me which equation is first second and thrid and reason why? i may be taking the wrong...
Applying Hess’s Law, from the enthalpies of reactions, N2(g) + 3H2(g) → 2 NH3(g) ΔH =...
Applying Hess’s Law, from the enthalpies of reactions, N2(g) + 3H2(g) → 2 NH3(g) ΔH = − 91.8 kJ O2(g) + 2H2(g) → 2H2O (g) ΔH = − 483.7 kJ N2(g) + O2(g) → 2NO(g) ΔH = 180.6 kJ Calculate the enthalpy change (ΔHrxn) for the reaction: 4NH3(g) + 5O2(g) → 4 NO (g) + 6H2O(g)
Calculate the standard enthalpy change, ΔH°rxn, in kJ for the following chemical equation, using only the...
Calculate the standard enthalpy change, ΔH°rxn, in kJ for the following chemical equation, using only the thermochemical equations below: 4KO2(s) + 2H2O(l) → 4KOH(aq) + 3O2(g) Report your answer to three significant figures in scientific notation. Equations:   ΔH°rxn (kJ) 4K(s) + O2(g) → 2K2O(s) -726.4 K(s) + O2(g) → KO2(s) -284.5 K2O(s) + H2O(l) → 2KOH(aq) -318
Using the information below determine the change in enthalpy for the following reaction:             2NO(g) +...
Using the information below determine the change in enthalpy for the following reaction:             2NO(g) + 5 H2(g) → 2NH3(g) + 2H2O(l) H2(g) + ½ O2(g) →   H2O(l)                     ∆H°= -285.8kJ N2(g) + O2(g)   → 2NO(g)                     ∆H°= +180.5 kJ 2NH3(g) → N2(g) + 3H2(g)                    ∆H°= + 92.22 kJ please explain step by step! thanks!
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT