Answer
1359K
Work Step by Step
Step 1: Let’s calculate the natural logarithm of equilibrium constant lnK at 1120K.
At equilibrium, $\Delta$$G_{R}$=$\Delta$$G_{R}^{o}$+RTlnK=0; so,
lnK=-$\frac{\Delta G_{R}^{o}}{RT}$=-$\frac{22000J/mol}{(8.31J/(mol·K))\times1120K}$$\approx$-2.36
Step 2: Van’s Hoff equation:
ln($K_2$)-ln($K_1$)=-$\frac{\Delta H_{R}^{o}}{R}$($\frac{1}{T_2}$-$\frac{1}{T_1}$).
Based on this equation, we can tell when standard reaction enthalpy $\Delta$$H_{R}^{o}$ is constant at a temperature range, the graph of lnK vs (1/T) is linear as shown by the image below. According to given conditions by the problem, this happens between 800K and 1500K.
In step 1 we got at T=1120K, lnK=-2.36. As T=1120K is between 800K and 1500K where $\Delta$$H_{R}^{o}$ is constant, we can blog it into Van’s Hoff equation to figure out lnK at T=800K and T=1500K:
lnK(T=800K)=ln(T=1120K)-$\frac{\Delta H_{R}^{o}}{R}$ ($\frac{1}{800K}$-$\frac{1}{1120K}$)
=-2.36-$\frac{125000J/mol}{8.31J/(mol·K)}$($\frac{1}{800K}$-$\frac{1}{1120K}$)
$\approx$-7.74
lnK(T=1500K)=ln(T=1120K)-$\frac{\Delta H_{R}^{o}}{R}$($\frac{1}{1500K}$-$\frac{1}{1120K}$)
=-2.36-$\frac{125000J/mol}{8.31J/(mol·K)}$($\frac{1}{1500K}$-$\frac{1}{1120K}$)
$\approx$+1.04
Step 3: When K>1 as the problem is asking, lnK>0. In step 2 we got lnK(T=800K)<0 and lnK(T=1500K)>0, so from 800K to 1500K, the graph of lnK vs (1/T) crosses the horizontal axis at the point where temperature T=x. When the temperature T>x (shown by the green region), lnK>0.
And we need figure out x where lnK=0. Let’s pick a temperature T from 800K to 1500K where $\Delta$$H_{R}^{o}$ is constant, let’s say 1120K; then blog it and x into Van’s Hoff equation:
lnK(T=x)=ln(T=1120K)-$\frac{\Delta H_{R}^{o}}{R}$($\frac{1}{x}$-$\frac{1}{1120K}$)=0
==> -2.36-$\frac{125000J/mol}{8.31J/(mol·K)}$($\frac{1}{x}$-$\frac{1}{1120K}$)=0
When solving for x, we get x$\approx$1359K.
