Question

calculate the work done in Joules when 4.5 gram H2O2 to react against a pressure of 1 ATM at 25 degree Celsius

Q15. Calculate the work done (in J) when 4.5 g H2O2 reacts against a pressure of 1 atm at 25°C

2H2O2(l) →O2(g) +H2O(l)

(A) – 1.63 × 102 (B) 4.5 × 102 (C) 3.2 × 102 (D) – 6.1 × 102

Answer 1

Answer : (A) The work done (in J) = -1.4976 × $10^{2}$ which is approximately equal to -1.63 × $10^{2}$.

Solution : Given,

Mass of $H_{2}O_{2}$ = 4.5 g

Pressure (P) = 1 atm

Temperature (T) = $25^{0}C$

Molar mass of $H_{2}O_{2}$ = 34 g/mole

Moles of $H_{2}O_{2}$ = $\frac{\text{ Mass of }H_{2}O_{2} }{\text{Molar mass of }H_{2}O_{2}}$ = $\frac{4.5g}{34g/mole}$ = 0.132 moles

The given reaction is,

$2H_{2}O_{2}(l)\rightarrow O_{2}(g)+2H_{2}O(l)$

Now, we will calculate the number of moles of $O_{2}$.

From the given reaction, we conclude that

2 moles of $H_{2}O_{2}$ gives             →     1 mole of $O_{2}$

0.132 moles of $H_{2}O_{2}$ gives      →      $\frac{1 mole\times 0.132 moles}{2 moles}$ of $O_{2}$

                                                            =      0.066 moles of $O_{2}$

We are taking volume of liquid as negligible. Therefore, we calculate the volume of $O_{2}$(g) and the initial volume will be equal to zero.

And final volume of $O_{2}$(g) during the ideal gas behavior = 0.066 × 22.4 L(at STP) = 1.4789 L

Formula used :

W = - P $(V_{final}-V_{initial})$

Total work done in negative because work done on system.

where,

w =  Total work done

P = pressure

$V_{final}$ = final volume

$V_{initiall}$ = initial volume

Now put all the values in above formula, we get

W = - 1 atm (1.4789 L - 0 L) = - 1.4789 L atm

Conversion :   1 L atm = 101.3 J

W = - 1.4789 L atm × 101.3 J/L atm = - 149.76 J = - 1.4976 × $10^{2}$ J $\approx$  -1.63 × $10^{2}$ J

Answer 2

Answer:The answer is as below

Explanation: