Question

sample of a gas in a cylindrical chamber with a movable piston occupied a volume of 6.414 liters when the pressure was 850 torr and the temperature was 27.2 C. The pressure was readjusted to 4423 torr by moving the piston. What was the volume occupied by the sample under the new conditions if the temperature remained constant throughout? PM = PM 26 A Ti T2 27. Determine the enthalpy change, Δ}, for the reaction, N2(g) + 2H2(g) → N2H4(/), given the following thermochemical equations: ΔH =-622.0 kJ Δ11=-285.9 kJ N2(g) N2H4(1) + O2(g) → 2H20(1) + H2(g) + ½ Oz(g)→ H2O(/)

Answer 1

26. You have to apply the equation

P1V1 / T1 = P2V2 / T2, p is pressure , t is temperature, V is volume, 1 and 2 refers to an initial and final state, since the temperature remains constant this equation is

P1* V1 = P2 * V2

850 * 6.414 = 4423 * V2

V2 = 850 * 6.414 / 4423 = 1.23 Liters

27.

Reverse the equation 1)

2 H2O + N2 ==== N2H4 + O2 H = 622 KJ, change the sign when you reverse an equation

multiply equation 2 by 2

2 H2 + O2 ===== 2H2O, -285.9 * 2 = -571.8 KJ

Add the equations

2 H2 + O2 ===== 2H2O

2 H2O + N2 ==== N2H4 + O2

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2H2 + N2 ======== N2H4 , our equations are right so just add the values we got

-571.8 + 622 = 50.2 KJ

Remember that if you reverse an equation you must change the sign from positive to negative or negative to positive

If you multiply an equation you have to multiply the value of the enthalpy

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