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NEET AIPMT SOLVED PAPER 2001

  • question_answer
                    Change in enthalpy for reaction \[2{{H}_{2}}{{O}_{2}}(l)\xrightarrow{{}}2{{H}_{2}}O(l)+{{O}_{2}}(g)\] If heat of formation of \[{{H}_{2}}{{O}_{2}}(l)\] and \[{{H}_{2}}O(l)\]are - 188 and -286 kJ/mol respectively:                                                                                            

    A)                 -196 k J/mol                       

    B)                 + 196 k J/mol

    C)                 + 948 k J/mol

    D)                 - 948 k J/mol

    Correct Answer: A

    Solution :

                    \[2{{H}_{2}}{{O}_{2}}(l)\xrightarrow{{}}2{{H}_{2}}{{O}_{2}}(l)+{{O}_{2}}(g)\Delta H=?\]                 \[\Delta H=[2\times \Delta {{H}_{f}}\,\text{of}\,{{H}_{2}}O(l)+(\Delta {{H}_{f}}\,\text{of}\,{{O}_{2}})]\]                                                                 \[-(2\times \Delta {{H}_{f}}\,\text{of}\,{{H}_{2}}{{O}_{2}}(l))]\]                 \[=[(2\times -286)+(0)-(2\times -188)]\]                 \[=[-572+376]\]                 \[=-196\,\,kJ\,/mol\]


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