ΔG deg will be negative above 7.27e+3 K.
<u>Explanation:</u>
- The ΔG deg with the temperature can be found using the formula and the formula is given below
- ΔG deg = ΔH deg - T ΔS deg
- Given data, ΔH deg = 181kJ and ΔSdeg=24.9J/K
- -T ΔS deg will be always negative and ΔG deg = ΔH deg will be positive and ΔG deg will be negative at relatively high temperatures and positive at relatively low temperatures
- solving the equation and substitute ΔGdeg=0
- ΔGdeg = ΔHdeg - T ΔSdeg
- T= ΔHdeg/ΔSdeg
- T=181 kJ / 2.49e-2 kJK-1
- By simplification we get
- T=7.27 × 10^3 K.
- Therefore, Go will be negative above 7.27 × 10^3 K
- Since ΔG deg = -RT lnK, when ΔGdeg < 0, K > 1 so the reaction will have K > 1 above 7.27 × 10^3 K.
- ΔG deg will be negative above 7.27e+3 K.
<u></u>
<u />
The first answer choice would be more conservative to our environment
Additional pressure required : 175 mm
<h3>Further explanation</h3>
Boyle's Law
At a constant temperature, the gas volume is inversely proportional to the pressure applied

V₁=250 ml
P₁=700 mmHg


Additional pressure :

A mixture<span> is the blending of two or more dissimilar substances. A major </span>characteristic<span> of </span>mixtures<span> is that the materials do not chemically combine. Hope this answers the question. Have a nice day. Feel free to ask more questions.</span>
Answer:
Activation energy, Ea, for the reaction 261.7 kj/mole
Explanation:
Given
Rate constant at 470 ⁰C (K₁) = 1.10 x 10⁻⁴ s⁻¹ and rate constant at 500⁰C (K₂) = 5.7 x 10⁻⁴ s⁻¹
Temperature (T₁) = 470 + 273 = 743 K and
temperature (T₂) = 500 + 273 = 773 K
Activation energy (Eₐ) = ?
Universal gas constant (R) = 8.314 J. K⁻¹. mole⁻¹
We know log
= 

⇒ log
=

⇒ 0.714 = 
⇒ Eₐ =
j/mole = 261.7 Kj/mole