Houston Texans quarterback C.J. Stroud has earned AFC Offensive Player of the Week honors after leading his team to a dramatic win over the Jacksonville Jaguars in London.
Stroud completed 24 of 34 passes for 312 yards and three touchdowns in Houston’s 34-31 Week 6 victory. Announced Wednesday, the award marks the fourth time in his career that he has received the conference honor.
He finished with a 116.3 passer rating and directed a seven-play, 36-yard drive late in the game. That series put Ka’imi Fairbairn in position to make the winning 49-yard field goal.
Stroud’s performance helped Houston earn its second consecutive win after an 0-4 start. The Texans improved to 2-4, while Jacksonville dropped to 4-2 after giving up the lead late in the fourth quarter.
The quarterback connected with tight end Foster Moreau for two touchdowns, while rookie receiver Jaylin Noel finished with 107 receiving yards and a score. Stroud’s production helped Houston prevail despite a late Jaguars rally.
Stroud’s 312 passing yards ranked second among AFC quarterbacks in Week 6, while his 116.3 passer rating was the highest in the NFL for the week. The performance gave Houston a timely boost in a divisional matchup.
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Houston now turns its attention to the New York Giants, who visit Sunday for a Week 7 matchup. The Texans will try to extend their winning streak before entering a Week 8 bye.
Stroud has completed 143 of 219 passes this season for 1,714 yards, nine touchdowns and one interception. He has thrown for at least 300 yards in three of six games, giving Houston a productive passing attack as it looks to climb the AFC South standings.
The honor recognizes his impact, but Houston needs consistency to turn this surge into a meaningful playoff push.
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