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Chemistry exam qn

  1. Jun 9, 2006 #1
    someone could juz tell me roughly how to work out this set of qn plz? i have been thinking and doing alot of workings but none of the ans is remotely close to the ans... :(

    1) the reaction below releases 56.6kj of heat at 298k for each mole of NO2 formed at a constant pressure of 1 atm. what is the standard enthalpy of formation of NO2 given the standard enthalpy of NO is 90.4kj mol

    2NO + O2 ---> 2NO2

    2) a 200g of copper at 100 degrees celsius is dropped into 1000g of water at 25 degrees celsius. what is the final temp of the system?

    specific heat of water is 4.18J and copper is 0.400 J

    3) if the equilibrium constant for A + B <===> C is 0.123, the equilibrium constant when 2C <===> 2A + 2B is?
     
  2. jcsd
  3. Jun 9, 2006 #2

    Hootenanny

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    Question Two

    HINT: Energy lost by copper is equal to the energy gained by the water. Try setting up simultaneous equations.

    Question Three

    HINT: You have increased the concentration of all the reactants equally.
     
  4. Jun 10, 2006 #3
    For Question One, use the fact that [tex]\Delta H = \Sigma (\Delta H_{products})-\Sigma (\Delta H_{reactants})[/tex]. You know that since oxygen is a pure element, its heat of formation is zero. You know that [tex]\Delta H[/tex], and you know [tex]\Sigma (\Delta H_{reactants})[/tex]. You have to find [tex]\Sigma (\Delta H_{products})[/tex].

    For Question Two, you know that [tex]q = mc(T_{f} - T_{i})[/tex]. Find the heat that the copper is holding. Now that you know [tex]q[/tex], you also know that it is all transferred to the water, so write another q-equation, but this time you are solving for T of the water. Find the equilibrium temperature of the water and copper system - that is your [tex]T_{f}[/tex] for the copper. You know the [tex]T_{i}[/tex] for both the copper and the water, so all you do now is plug and chug.

    For Question Three, write the [tex]K_{eq}[/tex] equation for A + B <===> C, then write it for 2C <===> 2A + 2B. Remember that when you flip the reactants and the products, you have to take the reciprocal of [tex]K_{eq}[/tex], and that when you multiply the coefficients all by a number [tex]N[/tex], you have to raise all of the terms in the [tex]K_{eq}[/tex] equation to that power [tex]N[/tex]. So, for example: The equilibrium constant for a reaction A + B + C <===> D + E + F is [tex]\frac{[D][E][F]}{[A][C]}[/tex]. For 3D + 3E + 3F <===> 3A + 3B + 3C, it is [tex]\frac{[A]^{3}^{3}[C]^{3}}{[D]^{3}[E]^{3}[F]^{3}}[/tex].
     
    Last edited: Jun 10, 2006
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