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Hess Law Practice Problems

Quiz by Amoi Salmon

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10 questions
Show answers
  • Q1
    Which of the following is an example of Hess Law?
    Calculating the enthalpy change of a reaction by adding the enthalpy changes of two or more reactions.
    Finding the molar mass of a substance by measuring its density.
    Using stoichiometry to calculate the amount of reactant needed in a reaction.
    Determining the equilibrium constant of a reaction by measuring the concentrations of reactants and products.
    30s
  • Q2
    What is the symbol for enthalpy change?
    ΔE
    ΔS
    ΔH
    ΔG
    30s
  • Q3
    What does a negative enthalpy change,  ΔH, indicate?
    That the reaction is exothermic.
    That the reaction does not involve a change in enthalpy.
    That the reaction is at equilibrium.
    That the reaction is endothermic.
    30s
  • Q4
    What is the enthalpy of reaction for the reverse of a reaction (flipped)?
    The negative of the enthalpy of the forward reaction.
    The same as the enthalpy of the forward reaction.
    Zero.
    It cannot be determined.
    30s
  • Q5
    What is the enthalpy of reaction for a reaction that has two or more reaction steps?
    The sum of the enthalpy changes of the individual reaction steps.
    It cannot be determined.
    Zero.
    The difference of the enthalpy changes of the individual reaction steps.
    30s
  • Q6

    A + B ----> 2C      ΔH = 50kJ

    D+ 2C----> E       ΔH = 25kJ

    What is the ΔH of the overall reaction: A+ B + D-----> E ?

    100kJ

    75kJ

    25kJ

    50kJ

    30s
  • Q7

    What is the ΔH of the overall reaction: 2A+ 5B ----> 2C + 4D

    A + 2B ---> C                 ΔH = -80kJ

    A + 3B ---> C + 4D     ΔH = -55kJ

    25kJ

    135kJ

    -25kJ

    -135kJ

    30s
  • Q8

    (3) Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: PCl3(l)  +  Cl2(g)→ PCl5(s)  

     2PCl3(l)  →  2P(s)  +  3Cl2(g)       ΔH = 640 kJ 

    2P(s)  +  5Cl2(g) →   2PCl5(s)      ΔH = -886 kJ

    123kJ

    -123kJ

    -246kJ

    246kJ

    30s
  • Q9

    Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: CuCl2(s) + Cu(s)  → 2CuCl(s)

    CuCl2(s)  →   Cu(s)  +  Cl2(g)             ΔH = 206 kJ

    2Cu(s) + Cl2(g)  →  2CuCl(s)           ΔH = -136 kJ

    342kJ

    70kJ

    -70kJ

    -342kJ

    30s
  • Q10

    What type of reaction is the following reaction and why?

    P4(s)  +  6Cl2(g)  →  4PCl3(g)            ΔH = -2439 kJ

    Exothermic because ΔH is positive

    Endothermic because ΔH is positive

    Endothermic because ΔH is negative

    Exothermic because ΔH is negative

    30s

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