Hess's Law

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The standard enthalpy of combustion of butane, in kJ mol⁻¹, is

• −2880

• −2590

• −806

• −554

Tags

NGSS.HS-PS1-4

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The value in kJ mol⁻¹ of the enthalpy of thermal dissociation when butane forms propane, hydrogen and carbon is

• −26.3

• −17.5

• +17.5

• +21.2

Tags

NGSS.HS-PS1-4

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The value in kJ mol−1 for the enthalpy of combustion of propane is

• −211.7

• −419.7

• −2220

• −2878

Tags

NGSS.HS-PS1-4

Using the data below, which is the correct value for the standard enthalpy of formation for TiCl₄(l)?

• −1538 kJ mol⁻¹

• −1094 kJ mol⁻¹

• −750 kJ mol⁻¹

• +286 kJ mol⁻¹

Tags

NGSS.HS-PS1-4

The enthalpies of combustion of C(s), H2(g) and C4H9OH(l) (in kJmol^-1) are as follows  
C(s) + O₂(g)   ->  CO₂(g)                                               ∆H=a
H₂(g) + ½O₂(g)   ->   H₂O(l)                                        ∆H=b
C₄H₉OH(l) + 6O₂(g)   ->   4CO₂(g) + 5H₂O(l)  ∆H=c
What is the enthalpy change for the reaction shown below?
  4C(g) + 5H₂(l) + ½O₂(g)   ->   C₄H₉OH(l)

• c – 4a – 5b
• 2a + 10b - c
• 4a + 5b - c
• 2a + 5b + c

Tags

NGSS.HS-PS1-4

(Hess's Law)Calculate the ∆H for the following reaction: 2H₂O₂ →  2H₂O  + 1 O₂ 
You are given these two equations:
2H₂  +  O₂ → 2H₂O            ∆H  =  -572 kJ
H₂  +  O₂  →  H₂O₂            ∆H  =  -188 kJ 

• ∆H  =  -948 kJ 
• ∆H  =  -196 kJ 
• ∆H  =  -384 kJ 
• ∆H  =  -188 kJ 

Tags

NGSS.HS-PS1-4

When you multiply a chemical equation by 2, what is done to the heat of reaction?

• multiplied by 1/2

• multiplied by 2

• nothing

Tags

NGSS.HS-PS1-4

When you flip a chemical reaction using Hess's Law, what is done to the heat of reaction value?

• nothing

• the sign is flipped

• 1/delta H

Tags

NGSS.HS-PS1-4