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Haryana PMT Haryana PMT Solved Paper-2007

  • question_answer
    For the reaction \[1g\,mole\]of \[CaC{{O}_{3}}\]is enclosed in 5 L container \[CaC{{O}_{3}}(s)\xrightarrow{{}}CaO(s)+C{{O}_{2}}(g)\]\[{{K}_{p}}=1.16\]at \[1073K\]then per cent dissociation of \[CaC{{O}_{3}}\]is

    A)  \[65%\]                              

    B)  \[100%\]

    C)  \[6.5%\]                             

    D)  zero

    Correct Answer: C

    Solution :

                    \[CaC{{O}_{3}}(s)\xrightarrow{{}}CaO(s)+C{{O}_{2}}(g)\] \[{{K}_{p}}={{P}_{C{{O}_{2}}}}\]    (only gaseous molecule count)                 \[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}\] \[\Rightarrow \]               \[1.16=\frac{x}{5}{{(0.0821\times 1073)}^{1}}\]                 \[x=\frac{1.16\times 5}{0.0821\times 1073}=\frac{5.8}{88.1}=0.0658\]                 \[=0.0658\times 100=6.58%\]              


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