1.

If 1/2X2O = X + 1/4O2 Change in H =90kJ Then heat change during reaction of metal X with one mole of O2 to form oxide to maximum extent (1) 360 kJ (2) -360 kJ (3) 180 kJ (4) -180 kJ

Answer»

ANSWER: The correct answer is OPTION (2).

Explanation:

Given reaction;

\frac{1}{2}X_2O\rightarrow <klux>X</klux>+\frac{1}{4}O_2,\Delta H=90kJ...(1)

On the reversing the reaction:

X+\frac{1}{4}O_2\rightarrow \frac{1}{2}X_2O,\Delta H=-90kJ..(2)

Then heat CHANGE during reaction of metal X with one mole of O_2 to form oxide to maximum extent.

Multiply the above the reaction (2) with '4'.

4X+O_2\rightarrow 2X_2O

So, the heat change during of the reaction = 4\times \Delta H=4\times (-90kJ)=-360kJ





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