C2H4(g) + 2O2(g) → 2CO(g) + 2H2O(g) ∆H1 = −758 kJ mol−1 ....1)
2C(s) + 2H2(g) → C2H4(g) ∆H2 = +52 kJ mol−1 ....2)
H2(g) + O2(g) → H2O(g) ∆H3 = −242 kJ mol−1 ....3)
Now, enthalpy of formation of carbon monoxide is given by :
∆H = ∆H1 + ∆H2 - ∆H3
∆H = ( -758 + 52 - ( -242 ) ) kJ mol−1
∆H = −464 kJ mol−1
Therefore, the enthalpy of formation of carbon monoxide is -464 kJ mol−1.
Hence, this is the required solution.
I would say D. Let me know if i am wrong.
Explanation:
It is given that two loads have 0.75 Ampere current each. And, they contain 2500 milli ampere per hour Ni-Cd battery.
As both the loads are connected in parallel. Hence, total current will be calculated as follows.
I = ![I_{1} + I_{2}](https://tex.z-dn.net/?f=I_%7B1%7D%20%2B%20I_%7B2%7D)
= 0.75 A + 0.75 A
= 1.5 A
= ![1.5 A \times \frac{1000 mA}{1 A}](https://tex.z-dn.net/?f=1.5%20A%20%5Ctimes%20%5Cfrac%7B1000%20mA%7D%7B1%20A%7D)
= 1500 mA
Relation between time and capacity of battery is as follows.
Capacity = Current × time (in hour)
therefore, time = ![\frac{Capacity}{Current}](https://tex.z-dn.net/?f=%5Cfrac%7BCapacity%7D%7BCurrent%7D)
= ![\frac{2500 mA. h}{1500 A}](https://tex.z-dn.net/?f=%5Cfrac%7B2500%20mA.%20h%7D%7B1500%20A%7D)
= 1.667 hr
Thus, we can conclude that the battery provide power to the load up to 1.667 hours.