Question

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 1

Answer: The correct answer is option (2).

Explanation:

Given reaction;

$\frac{1}{2}X_2O\rightarrow X+\frac{1}{4}O_2,\Delta H=90kJ$...(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 = $4\times \Delta H=4\times (-90kJ)=-360kJ$

Answer 2

Answer:

The answer for this question is -360KJ